Further Calculus

Further Calculus

This topic extends the calculus of A Level Mathematics to more powerful integration techniques, inverse trigonometric functions, volumes of revolution, arc length, and surface area. These tools are essential for university-level mathematics, physics, and engineering.

Board Coverage

Board	Paper	Notes
AQA	Paper 1	Integration by parts, inverse trig integrals, volumes, arc length
Edexcel	FP1, FP2	Parts in FP1; inverse trig, volumes, arc length in FP2
OCR (A)	Paper 1	Parts, inverse trig integrals, volumes
CIE (9231)	P1, P2	Parts and volumes in P1; arc length and surface area in P2

All boards provide standard integrals in the formula booklet. You must know how to apply

integration by parts repeatedly, derive and use reduction formulae, and set up volumes of revolution integrals correctly. CIE places particular emphasis on parametric volumes of revolution. :::

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1. Integration by Parts (Advanced)

1.1 The formula — proof from the product rule

Theorem. For differentiable functions $u(x)$ and $v(x)$:

$\boxed{\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx}$

Proof of the integration by parts formula

From the product rule:

$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

Integrating both sides with respect to $x$:

$uv = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx$

Rearranging:

$\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx \quad \blacksquare$

1.2 Repeated integration by parts

When the integral does not simplify in one step, apply integration by parts repeatedly until it does.

Example. Find $\displaystyle\int x^2 e^x\,dx$.

First application: $u = x^2$, $dv = e^x\,dx$. $du = 2x\,dx$, $v = e^x$.

$\int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx$

Second application on $\int x e^x\,dx$: $u = x$, $dv = e^x\,dx$. $du = dx$, $v = e^x$.

$\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C$

Therefore:

$\int x^2 e^x\,dx = x^2 e^x - 2(xe^x - e^x) + C = e^x(x^2 - 2x + 2) + C$

Example. Find $\displaystyle\int e^{ax}\cos bx\,dx$.

Let $I = \int e^{ax}\cos bx\,dx$. First application: $u = e^{ax}$, $dv = \cos bx\,dx$.

$du = ae^{ax}\,dx, \quad v = \frac{1}{b}\sin bx$

$I = \frac◆LB◆e^{ax}\sin bx◆RB◆◆LB◆b◆RB◆ - \frac{a}{b}\int e^{ax}\sin bx\,dx$

Second application on $\int e^{ax}\sin bx\,dx$: $u = e^{ax}$, $dv = \sin bx\,dx$.

$du = ae^{ax}\,dx, \quad v = -\frac{1}{b}\cos bx$

$\int e^{ax}\sin bx\,dx = -\frac◆LB◆e^{ax}\cos bx◆RB◆◆LB◆b◆RB◆ + \frac{a}{b}\int e^{ax}\cos bx\,dx = -\frac◆LB◆e^{ax}\cos bx◆RB◆◆LB◆b◆RB◆ + \frac{a}{b}I$

Substituting back:

$I = \frac◆LB◆e^{ax}\sin bx◆RB◆◆LB◆b◆RB◆ - \frac{a}{b}\left(-\frac◆LB◆e^{ax}\cos bx◆RB◆◆LB◆b◆RB◆ + \frac{a}{b}I\right)$

$I = \frac◆LB◆e^{ax}\sin bx◆RB◆◆LB◆b◆RB◆ + \frac◆LB◆ae^{ax}\cos bx◆RB◆◆LB◆b^2◆RB◆ - \frac{a^2}{b^2}I$

$I\left(1 + \frac{a^2}{b^2}\right) = e^{ax}\left(\frac◆LB◆\sin bx◆RB◆◆LB◆b◆RB◆ + \frac◆LB◆a\cos bx◆RB◆◆LB◆b^2◆RB◆\right)$

$\boxed{I = \frac◆LB◆e^{ax}(a\cos bx + b\sin bx)◆RB◆◆LB◆a^2 + b^2◆RB◆ + C}$

tip original integral reappears — solve for it algebraically. Always keep $u = e^{ax}$ on

both applications. :::

1.3 Reduction formulae

A reduction formula expresses an integral $I_n$ (depending on a parameter $n$) in terms of $I_{n-1}$ or $I_{n-2}$.

Proof of the reduction formula for $I_n = \int_0^{\pi/2} \sin^n x\,dx$

Write $I_n = \int_0^{\pi/2}\sin^{n-1}x \cdot \sin x\,dx$.

Let $u = \sin^{n-1}x$, $dv = \sin x\,dx$. Then:

$du = (n-1)\sin^{n-2}x\cos x\,dx, \quad v = -\cos x$

$I_n = \bigl[-\sin^{n-1}x\cos x\bigr]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx$

The boundary term vanishes: at $x = 0$, $\sin 0 = 0$; at $x = \pi/2$, $\cos(\pi/2) = 0$.

Using $\cos^2 x = 1 - \sin^2 x$:

$I_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x\,dx - (n-1)\int_0^{\pi/2}\sin^n x\,dx$

$I_n = (n-1)I_{n-2} - (n-1)I_n$

$nI_n = (n-1)I_{n-2}$

$\boxed{I_n = \frac{n-1}{n}\,I_{n-2}, \quad n \geq 2}$

The base cases are $I_0 = \displaystyle\int_0^{\pi/2}1\,dx = \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$ and $I_1 = \displaystyle\int_0^{\pi/2}\sin x\,dx = 1$.

Example. Using the reduction formula, $I_4 = \dfrac{3}{4}I_2 = \dfrac{3}{4}\cdot\dfrac{1}{2}I_0 = \dfrac{3}{4}\cdot\dfrac{1}{2}\cdot\dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ = \dfrac◆LB◆3\pi◆RB◆◆LB◆16◆RB◆$.

warning $0$ and $\pi/2$. For general limits, the boundary term must be evaluated. :::

Example. Find a reduction formula for $I_n = \displaystyle\int x^n e^x\,dx$.

Let $u = x^n$, $dv = e^x\,dx$. Then $du = nx^{n-1}\,dx$, $v = e^x$.

$\boxed{I_n = x^n e^x - nI_{n-1}}$

With $I_0 = e^x + C$.

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2. Integration Using Partial Fractions

2.1 Linear factors

When the denominator factorises into distinct linear factors, decompose and integrate each term.

Example. $\displaystyle\int \frac{2x+3}{(x+1)(x+2)}\,dx$.

$\frac{2x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$

$2x + 3 = A(x+2) + B(x+1)$. Setting $x = -1$: $1 = A$. Setting $x = -2$: $-1 = -B \implies B = 1$.

$\int \frac{1}{x+1} + \frac{1}{x+2}\,dx = \ln|x+1| + \ln|x+2| + C = \ln|(x+1)(x+2)| + C$

2.2 Irreducible quadratic factors

When the denominator contains an irreducible quadratic $ax^2 + bx + c$ (discriminant $b^2 - 4ac < 0$), the partial fraction has a linear numerator over the quadratic, leading to $\ln$ and $\arctan$ terms.

Example. $\displaystyle\int \frac{3x + 1}{x^2 + 2x + 5}\,dx$.

Complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$.

Split the numerator to match the derivative of the denominator:

$\frac{3x+1}{x^2+2x+5} = \frac◆LB◆\frac{3}{2}(2x+2) + 1 - 3◆RB◆◆LB◆x^2+2x+5◆RB◆ = \frac{3}{2}\cdot\frac{2x+2}{x^2+2x+5} - \frac{2}{(x+1)^2+4}$

$\int \frac{3x+1}{x^2+2x+5}\,dx = \frac{3}{2}\ln(x^2+2x+5) - 2\cdot\frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C$

$= \frac{3}{2}\ln(x^2+2x+5) - \arctan\!\left(\frac{x+1}{2}\right) + C$

tip into a multiple of $(2x + b)$ plus a constant, (3) the $(2x+b)$ part gives $\ln$, the

constant gives $\arctan$. :::

2.3 Repeated factors

Example. $\displaystyle\int \frac{1}{x(x-1)^2}\,dx$.

$\frac{1}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$

$1 = A(x-1)^2 + Bx(x-1) + Cx$.

$x = 0$: $1 = A$. $x = 1$: $1 = C$. $x = 2$: $1 = A + 2B + 2C = 1 + 2B + 2 \implies B = -1$.

$\int\left(\frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}\right)dx = \ln|x| - \ln|x-1| - \frac{1}{x-1} + C$

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3. Inverse Trigonometric Integration

3.1 Derivation of the standard integrals

Theorem. The following integrals hold for $a > 0$:

$\boxed{\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C}$

$\boxed{\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{a^2-x^2}◆RB◆\,dx = \arcsin\frac{x}{a} + C}$

$\boxed{\int \frac{1}{a^2-x^2}\,dx = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C}$

Proof of $\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C$

Let $x = a\tan\theta$, so $dx = a\sec^2\theta\,d\theta$.

$\int \frac◆LB◆1◆RB◆◆LB◆a^2 + a^2\tan^2\theta◆RB◆\cdot a\sec^2\theta\,d\theta = \int \frac◆LB◆a\sec^2\theta◆RB◆◆LB◆a^2\sec^2\theta◆RB◆\,d\theta = \frac{1}{a}\int 1\,d\theta = \frac◆LB◆\theta◆RB◆◆LB◆a◆RB◆ + C$

Since $\theta = \arctan(x/a)$:

$\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C \quad \blacksquare$

Proof of $\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{a^2-x^2}◆RB◆\,dx = \arcsin\frac{x}{a} + C$

Let $x = a\sin\theta$, so $dx = a\cos\theta\,d\theta$ and $\sqrt{a^2 - x^2} = a\cos\theta$ (for $|\theta| \leq \pi/2$).

$\int \frac◆LB◆a\cos\theta◆RB◆◆LB◆a\cos\theta◆RB◆\,d\theta = \int 1\,d\theta = \theta + C = \arcsin\frac{x}{a} + C \quad \blacksquare$

Proof of $\int \frac{1}{a^2-x^2}\,dx = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C$

By partial fractions:

$\frac{1}{a^2-x^2} = \frac{1}{(a-x)(a+x)} = \frac{1}{2a}\left(\frac{1}{a-x} + \frac{1}{a+x}\right)$

$\int \frac{1}{a^2-x^2}\,dx = \frac{1}{2a}\bigl[-\ln|a-x| + \ln|a+x|\bigr] + C = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C \quad \blacksquare$

3.2 More general forms

$\int \frac{1}{a^2 + (x+b)^2}\,dx = \frac{1}{a}\arctan\frac{x+b}{a} + C$

$\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{a^2 - (x+b)^2}◆RB◆\,dx = \arcsin\frac{x+b}{a} + C$

These follow directly from the standard forms via the substitution $u = x + b$.

warning $\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{a^2-x^2}◆RB◆$ (gives $\arcsin$), and

$\dfrac{1}{a^2-x^2}$ (gives a logarithmic form). The square root makes the difference. :::

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4. Differentiation of Inverse Trigonometric Functions

4.1 Derivatives

$\boxed{\frac{d}{dx}\arcsin x = \frac◆LB◆1◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆, \quad |x| < 1}$

$\boxed{\frac{d}{dx}\arccos x = -\frac◆LB◆1◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆, \quad |x| < 1}$

$\boxed{\frac{d}{dx}\arctan x = \frac{1}{1+x^2}}$

Proof of $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$

Let $y = \arctan x$. Then $x = \tan y$.

Differentiating implicitly with respect to $x$:

$1 = \sec^2 y \cdot \frac{dy}{dx}$

$\frac{dy}{dx} = \frac◆LB◆1◆RB◆◆LB◆\sec^2 y◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆1 + \tan^2 y◆RB◆ = \frac{1}{1+x^2} \quad \blacksquare$

Proof of $\frac{d}{dx}\arcsin x = \frac◆LB◆1◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆$

Let $y = \arcsin x$. Then $x = \sin y$.

Differentiating implicitly:

$1 = \cos y \cdot \frac{dy}{dx}$

Since $\arcsin x$ has range $[-\pi/2, \pi/2]$, we have $\cos y \geq 0$, so $\cos y = \sqrt◆LB◆1-\sin^2 y◆RB◆ = \sqrt{1-x^2}$.

$\frac{dy}{dx} = \frac◆LB◆1◆RB◆◆LB◆\cos y◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆ \quad \blacksquare$

4.2 Chain rule with inverse trig functions

Example. $\dfrac{d}{dx}\arcsin(3x) = \dfrac◆LB◆3◆RB◆◆LB◆\sqrt{1-9x^2}◆RB◆$.

Example. $\dfrac{d}{dx}\arctan\!\left(\dfrac{x}{2}\right) = \dfrac{1/2}{1 + x^2/4} = \dfrac{2}{4+x^2}$.

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5. Volumes of Revolution

5.1 Rotation about the $x$-axis

Definition. The volume generated by rotating the region bounded by $y = f(x)$, the $x$-axis, $x = a$, and $x = b$ about the $x$-axis is:

$\boxed{V = \pi\int_a^b y^2\,dx = \pi\int_a^b [f(x)]^2\,dx}$

5.2 Rotation about the $y$-axis

Definition. The volume generated by rotating the region bounded by $x = g(y)$, the $y$-axis, $y = c$, and $y = d$ about the $y$-axis is:

$\boxed{V = \pi\int_c^d x^2\,dy = \pi\int_c^d [g(y)]^2\,dy}$

5.3 Parametric curves

When a curve is given parametrically by $x = x(t)$, $y = y(t)$:

• Rotation about the $x$-axis: $V = \pi\displaystyle\int_{t_1}^{t_2} y^2\,\frac{dx}{dt}\,dt$
• Rotation about the $y$-axis: $V = \pi\displaystyle\int_{t_1}^{t_2} x^2\,\frac{dy}{dt}\,dt$

The parametric volume formula uses $\dfrac{dx}{dt}$ or $\dfrac{dy}{dt}$ as appropriate.

Do not forget this factor — it is a very common error. :::

Example. Find the volume generated by rotating the curve $y = \sqrt{x}$ from $x = 0$ to $x = 4$ about the $x$-axis.

$V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$

Example. The curve $x = 2\cos t$, $y = 2\sin t$ for $0 \leq t \leq \pi$ is rotated about the $x$-axis. Find the volume.

$V = \pi\int_0^{\pi} (2\sin t)^2 \cdot \frac{dx}{dt}\,dt = \pi\int_0^{\pi} 4\sin^2 t \cdot (-2\sin t)\,dt$

$= -8\pi\int_0^{\pi}\sin^3 t\,dt = 8\pi\int_0^{\pi}\sin^3 t\,dt$

Using $\sin^3 t = \sin t(1-\cos^2 t)$ and the substitution $u = \cos t$:

$= 8\pi\int_{-1}^{1}(1-u^2)\,du = 8\pi\left[u - \frac{u^3}{3}\right]_{-1}^1 = 8\pi\left(\frac{2}{3} - \left(-\frac{2}{3}\right)\right) = \frac◆LB◆32\pi◆RB◆◆LB◆3◆RB◆$

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6. Arc Length and Surface Area of Revolution

6.1 Arc length

Theorem. For a curve $y = f(x)$ from $x = a$ to $x = b$:

$\boxed{s = \int_a^b \sqrt◆LB◆1 + \left(\frac{dy}{dx}\right)^2◆RB◆\,dx}$

For a curve given parametrically by $x = x(t)$, $y = y(t)$ from $t = t_1$ to $t = t_2$:

$\boxed{s = \int_{t_1}^{t_2} \sqrt◆LB◆\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2◆RB◆\,dt}$

Derivation (Cartesian). The arc length element $ds$ satisfies $ds^2 = dx^2 + dy^2$ by the Pythagorean theorem applied to an infinitesimal segment. Therefore:

$ds = \sqrt◆LB◆1 + \left(\frac{dy}{dx}\right)^2◆RB◆\,dx$

Integrating from $a$ to $b$ gives the total arc length.

Example. Find the arc length of $y = \ln(\cos x)$ from $x = 0$ to $x = \pi/3$.

$\frac{dy}{dx} = \frac◆LB◆-\sin x◆RB◆◆LB◆\cos x◆RB◆ = -\tan x$

$s = \int_0^{\pi/3}\sqrt◆LB◆1+\tan^2 x◆RB◆\,dx = \int_0^{\pi/3}\sec x\,dx = \Bigl[\ln|\sec x + \tan x|\Bigr]_0^{\pi/3}$

$= \ln(2 + \sqrt{3}) - \ln(1) = \ln(2+\sqrt{3})$

6.2 Surface area of revolution

Theorem. The surface area generated by rotating $y = f(x)$ from $x = a$ to $x = b$ about the $x$-axis:

$\boxed{S = 2\pi\int_a^b y\,\sqrt◆LB◆1 + \left(\frac{dy}{dx}\right)^2◆RB◆\,dx}$

For a parametric curve rotated about the $x$-axis:

$\boxed{S = 2\pi\int_{t_1}^{t_2} y\,\sqrt◆LB◆\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2◆RB◆\,dt}$

Example. Find the surface area generated by rotating $y = x^2$ from $x = 0$ to $x = 1$ about the $x$-axis.

$S = 2\pi\int_0^1 x^2\sqrt{1+4x^2}\,dx$

Let $x = \frac{1}{2}\tan\theta$, $dx = \frac{1}{2}\sec^2\theta\,d\theta$. When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \arctan 2$.

$$S = 2\pi\int_0^{\arctan 2}\frac◆LB◆\tan^2\theta◆RB◆◆LB◆4◆RB◆\cdot\sec\theta\cdot\frac{1}{2}\sec^2\theta\,d\theta = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\int_0^{\arctan 2}\tan^2\theta\sec^3\theta\,d\theta$$

Using $\tan^2\theta = \sec^2\theta - 1$ and integrating by parts with $u = \sec\theta$, $dv = \sec^2\theta\,d\theta$:

This integral evaluates to $\dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\left[\dfrac{1}{4}\sec\theta\tan\theta + \dfrac{1}{4}\ln|\sec\theta+\tan\theta| - \dfrac{1}{4}\sec\theta\tan\theta + \dfrac{1}{8}\ln|\sec\theta+\tan\theta|\right]_0^{\arctan 2}$.

Simplifying with $\sec(\arctan 2) = \sqrt{5}$ and $\tan(\arctan 2) = 2$:

$S = \frac◆LB◆9\pi\sqrt{5}◆RB◆◆LB◆16◆RB◆ - \frac◆LB◆\pi◆RB◆◆LB◆32◆RB◆\ln(2+\sqrt{5})$

CIE (9231) P2 requires arc length and surface area of revolution. Edexcel FP2 covers arc

length but surface area appears less frequently. AQA covers both in Paper 1. OCR (A) covers arc length in Paper 1. :::

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7. Summary of Key Results

Integral	Result
$\displaystyle\int\frac{1}{a^2+x^2}\,dx$	$\dfrac{1}{a}\arctan\dfrac{x}{a}+C$
$\displaystyle\int\frac◆LB◆1◆RB◆◆LB◆\sqrt{a^2-x^2}◆RB◆\,dx$	$\arcsin\dfrac{x}{a}+C$
$\displaystyle\int\frac{1}{a^2-x^2}\,dx$	$\dfrac{1}{2a}\ln\left\|\dfrac{a+x}{a-x}\right\|+C$
$\dfrac{d}{dx}\arcsin x$	$\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆$
$\dfrac{d}{dx}\arctan x$	$\dfrac{1}{1+x^2}$
Vol. about $x$-axis	$\pi\displaystyle\int_a^b y^2\,dx$
Arc length	$\displaystyle\int\sqrt◆LB◆1+\left(\frac{dy}{dx}\right)^2◆RB◆\,dx$

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Problems

Details

Problem 1

Find $\displaystyle\int x^3 e^{-x}\,dx$.

Details

Hint 1

Apply integration by parts three times, reducing the power of $x$ each time.

Details

Answer 1

First: $u = x^3$, $dv = e^{-x}\,dx$. $du = 3x^2\,dx$, $v = -e^{-x}$.

$\displaystyle\int x^3 e^{-x}\,dx = -x^3 e^{-x} + 3\int x^2 e^{-x}\,dx$

Second: $u = x^2$, $dv = e^{-x}\,dx$. $\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int x e^{-x}\,dx$.

Third: $u = x$, $dv = e^{-x}\,dx$. $\int x e^{-x}\,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x} - e^{-x}$.

Combining: $\int x^3 e^{-x}\,dx = -x^3 e^{-x} - 3x^2 e^{-x} - 6x e^{-x} - 6e^{-x} + C = -e^{-x}(x^3 + 3x^2 + 6x + 6) + C$.

Details

Problem 2

Find a reduction formula for $I_n = \displaystyle\int \cos^n x\,dx$ for $n \geq 2$.

Details

Hint 2

Write $\cos^n x = \cos^{n-1}x \cdot \cos x$ and apply integration by parts with $u = \cos^{n-1}x$, $dv = \cos x\,dx$. Use $\sin^2 x = 1 - \cos^2 x$.

Details

Answer 2

$I_n = \int\cos^{n-1}x\cos x\,dx$. Let $u = \cos^{n-1}x$, $dv = \cos x\,dx$.

$du = -(n-1)\cos^{n-2}x\sin x\,dx$, $v = \sin x$.

$I_n = \cos^{n-1}x\sin x + (n-1)\int\cos^{n-2}x\sin^2 x\,dx$

$= \cos^{n-1}x\sin x + (n-1)\int\cos^{n-2}x(1-\cos^2 x)\,dx$

$= \cos^{n-1}x\sin x + (n-1)I_{n-2} - (n-1)I_n$

$nI_n = \cos^{n-1}x\sin x + (n-1)I_{n-2}$

$\boxed{I_n = \frac{1}{n}\cos^{n-1}x\sin x + \frac{n-1}{n}I_{n-2} + C}$

Details

Problem 3

Evaluate $\displaystyle\int \frac{2x-1}{x^2+4x+13}\,dx$.

Details

Hint 3

Complete the square: $x^2+4x+13 = (x+2)^2+9$. Split the numerator into a multiple of $(2x+4)$ plus a constant.

Details

Answer 3

$x^2+4x+13 = (x+2)^2+9$. Write $2x-1 = (2x+4) - 5$.

$\displaystyle\int\frac{2x+4}{(x+2)^2+9}\,dx - 5\int\frac{1}{(x+2)^2+9}\,dx$

$= \ln(x^2+4x+13) - \frac{5}{3}\arctan\!\left(\frac{x+2}{3}\right) + C$

Details

Problem 4

Evaluate $\displaystyle\int_0^{\pi/2} \sin^5 x\,dx$ using the reduction formula.

Details

Hint 4

Use $I_n = \dfrac{n-1}{n}I_{n-2}$ with $I_1 = 1$.

Details

Answer 4

$I_5 = \dfrac{4}{5}I_3 = \dfrac{4}{5}\cdot\dfrac{2}{3}I_1 = \dfrac{4}{5}\cdot\dfrac{2}{3}\cdot 1 = \dfrac{8}{15}$.

Details

Problem 5

Find the volume generated by rotating the region bounded by $y = x^2$, $y = 0$, $x = 0$, $x = 2$ about the $y$-axis.

Details

Hint 5

Express $x$ in terms of $y$ ($x = \sqrt{y}$) and integrate with respect to $y$ from $0$ to $4$, or use the shell method: $V = 2\pi\int_0^2 x\cdot x^2\,dx$.

Details

Answer 5

Using the disc method about the $y$-axis: $V = \pi\int_0^4 x^2\,dy = \pi\int_0^4 y\,dy = \pi\left[\frac{y^2}{2}\right]_0^4 = 8\pi$.

Details

Problem 6

Find $\dfrac{d}{dx}\left[x\arcsin x + \sqrt{1-x^2}\right]$ and hence evaluate $\displaystyle\int_0^{1/2} \arcsin x\,dx$.

Details

Hint 6

Differentiate using the product rule and the chain rule with $\dfrac{d}{dx}\arcsin x$.

Details

Answer 6

$\dfrac{d}{dx}\bigl[x\arcsin x + \sqrt{1-x^2}\bigr] = \arcsin x + \dfrac◆LB◆x◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆ + \dfrac◆LB◆-x◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆ = \arcsin x$.

Therefore $\displaystyle\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C$.

$\displaystyle\int_0^{1/2}\arcsin x\,dx = \left[x\arcsin x + \sqrt{1-x^2}\right]_0^{1/2} = \frac{1}{2}\cdot\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ + \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ - 1 = \frac◆LB◆\pi◆RB◆◆LB◆12◆RB◆ + \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ - 1$.

Details

Problem 7

Find the arc length of the curve $y = \dfrac{x^3}{6} + \dfrac{1}{2x}$ from $x = 1$ to $x = 3$.

Details

Hint 7

Compute $\dfrac{dy}{dx} = \dfrac{x^2}{2} - \dfrac{1}{2x^2}$. Show that $1+\left(\dfrac{dy}{dx}\right)^2$ is a perfect square.

Details

Answer 7

$\dfrac{dy}{dx} = \dfrac{x^2}{2} - \dfrac{1}{2x^2}$.

$1+\left(\dfrac{dy}{dx}\right)^2 = 1 + \dfrac{x^4}{4} - \dfrac{1}{2} + \dfrac{1}{4x^4} = \dfrac{x^4}{4} + \dfrac{1}{2} + \dfrac{1}{4x^4} = \left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)^2$

$s = \displaystyle\int_1^3\left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)dx = \left[\dfrac{x^3}{6} - \dfrac{1}{2x}\right]_1^3 = \left(\dfrac{9}{2}-\dfrac{1}{6}\right)-\left(\dfrac{1}{6}-\dfrac{1}{2}\right) = \dfrac{14}{3}$.

Details

Problem 8

Evaluate $\displaystyle\int_0^{\infty}\frac{1}{4+x^2}\,dx$.

Details

Hint 8

Use $\displaystyle\int\frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a}$. Here $a = 2$.

Details

Answer 8

$\displaystyle\int_0^{\infty}\frac{1}{4+x^2}\,dx = \left[\frac{1}{2}\arctan\frac{x}{2}\right]_0^{\infty} = \frac{1}{2}\cdot\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - 0 = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$.

Details

Problem 9

The curve $x = t^2$, $y = t^3$ for $0 \leq t \leq 2$ is rotated about the $x$-axis. Find the volume of revolution.

Details

Hint 9

Use $V = \pi\displaystyle\int_{t_1}^{t_2}y^2\,\frac{dx}{dt}\,dt$. Here $\dfrac{dx}{dt} = 2t$.

Details

Answer 9

$V = \pi\displaystyle\int_0^2 t^6 \cdot 2t\,dt = 2\pi\int_0^2 t^7\,dt = 2\pi\left[\frac{t^8}{8}\right]_0^2 = 2\pi\cdot\frac{256}{8} = 64\pi$.

Details

Problem 10

Find $\displaystyle\int e^x\sin 2x\,dx$.

Details

Hint 10

Apply integration by parts twice. Keep $u = e^x$ on both applications. The original integral will reappear.

Details

Answer 10

Let $I = \displaystyle\int e^x\sin 2x\,dx$. First: $u = e^x$, $dv = \sin 2x\,dx$. $du = e^x\,dx$, $v = -\frac{1}{2}\cos 2x$.

$I = -\frac{1}{2}e^x\cos 2x + \frac{1}{2}\int e^x\cos 2x\,dx$.

Second on $\int e^x\cos 2x\,dx$: $u = e^x$, $dv = \cos 2x\,dx$. $du = e^x\,dx$, $v = \frac{1}{2}\sin 2x$.

$\int e^x\cos 2x\,dx = \frac{1}{2}e^x\sin 2x - \frac{1}{2}\int e^x\sin 2x\,dx = \frac{1}{2}e^x\sin 2x - \frac{1}{2}I$.

$I = -\frac{1}{2}e^x\cos 2x + \frac{1}{4}e^x\sin 2x - \frac{1}{4}I$.

$\frac{5}{4}I = e^x\left(\frac◆LB◆\sin 2x◆RB◆◆LB◆4◆RB◆ - \frac◆LB◆\cos 2x◆RB◆◆LB◆2◆RB◆\right)$.

$\boxed{I = \frac◆LB◆e^x(\sin 2x - 2\cos 2x)◆RB◆◆LB◆5◆RB◆ + C}$

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8. Advanced Worked Examples

Example 8.1: Leibniz's rule for higher derivatives of a product

Problem. If $y = x^2 e^{3x}$, find $\dfrac{d^4 y}{dx^4}$.

Solution. We use Leibniz's rule: $(uv)^{(n)} = \displaystyle\sum_{k=0}^{n} \binom{n}{k} u^{(k)} v^{(n-k)}$.

Let $u = x^2$ and $v = e^{3x}$.

• $u' = 2x$, $u'' = 2$, $u''' = 0$, $u^{(4)} = 0$.
• $v^{(k)} = 3^k e^{3x}$ for all $k$.

$\frac{d^4 y}{dx^4} = \binom{4}{0} x^2 \cdot 3^4 e^{3x} + \binom{4}{1} 2x \cdot 3^3 e^{3x} + \binom{4}{2} 2 \cdot 3^2 e^{3x} + 0 + 0$

$= 81x^2 e^{3x} + 4 \cdot 54x e^{3x} + 6 \cdot 18 e^{3x}$

$\boxed{= (81x^2 + 216x + 108)e^{3x}}$

Example 8.2: Reduction formula for $\int x^n e^x\,dx$

Problem. Establish and use a reduction formula for $I_n = \int x^n e^x\,dx$.

Solution. Using integration by parts with $u = x^n$, $dv = e^x\,dx$:

$I_n = x^n e^x - \int nx^{n-1} e^x\,dx = x^n e^x - nI_{n-1}$

Therefore $\boxed{I_n = x^n e^x - nI_{n-1}}$ with $I_0 = e^x + C$.

To find $I_3$:

$I_1 = x e^x - e^x$, $I_2 = x^2 e^x - 2x e^x + 2e^x$, $I_3 = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x$.

$\boxed{I_3 = (x^3 - 3x^2 + 6x - 6)e^x + C}$

Example 8.3: Improper integral convergence test

Problem. Determine whether $\displaystyle\int_0^1 \frac◆LB◆1◆RB◆◆LB◆\sqrt{x}◆RB◆\,dx$ converges, and evaluate if it does.

Solution. The integrand is undefined at $x = 0$. Write:

$\int_0^1 x^{-1/2}\,dx = \lim_{a \to 0^+} \int_a^1 x^{-1/2}\,dx = \lim_{a \to 0^+} \left[2x^{1/2}\right]_a^1 = \lim_{a \to 0^+} (2 - 2\sqrt{a}) = 2$

Since the limit exists and is finite, the integral converges. $\boxed{\displaystyle\int_0^1 \frac◆LB◆1◆RB◆◆LB◆\sqrt{x}◆RB◆\,dx = 2}$

Example 8.4: Integration using the $t = \tan(x/2)$ substitution

Problem. Evaluate $\displaystyle\int_0^{\pi/2} \frac◆LB◆1◆RB◆◆LB◆1 + \sin x◆RB◆\,dx$ using the Weierstrass substitution.

Solution. Let $t = \tan(x/2)$, so $\sin x = \dfrac{2t}{1+t^2}$ and $dx = \dfrac{2\,dt}{1+t^2}$.

When $x = 0$: $t = 0$. When $x = \pi/2$: $t = 1$.

$\int_0^1 \frac◆LB◆1◆RB◆◆LB◆1 + \frac{2t}{1+t^2}◆RB◆ \cdot \frac{2\,dt}{1+t^2} = \int_0^1 \frac{2\,dt}{(1+t^2) + 2t} = \int_0^1 \frac{2\,dt}{t^2 + 2t + 1} = \int_0^1 \frac{2\,dt}{(t+1)^2}$

$= \left[-\frac{2}{t+1}\right]_0^1 = -1 + 2 = \boxed{1}$

Example 8.5: Differentiation of parametric arc length

Problem. A curve is given by $x = t - \sin t$, $y = 1 - \cos t$ for $0 \leq t \leq 2\pi$. Find the total arc length.

Solution. $\dfrac{dx}{dt} = 1 - \cos t$, $\dfrac{dy}{dt} = \sin t$.

$s = \int_0^{2\pi} \sqrt◆LB◆(1-\cos t)^2 + \sin^2 t◆RB◆\,dt = \int_0^{2\pi} \sqrt◆LB◆1 - 2\cos t + \cos^2 t + \sin^2 t◆RB◆\,dt$

$= \int_0^{2\pi} \sqrt◆LB◆2 - 2\cos t◆RB◆\,dt = \int_0^{2\pi} \sqrt◆LB◆4\sin^2(t/2)◆RB◆\,dt = \int_0^{2\pi} 2|\sin(t/2)|\,dt$

For $0 \leq t \leq 2\pi$, $\sin(t/2) \geq 0$, so:

$s = 2\int_0^{2\pi} \sin(t/2)\,dt = 2\left[-2\cos(t/2)\right]_0^{2\pi} = 2(2 + 2) = \boxed{8}$

Example 8.6: Taylor series approach to a difficult limit

Problem. Evaluate $\displaystyle\lim_{x \to 0} \frac◆LB◆x - \sin x◆RB◆◆LB◆x^3◆RB◆$.

Solution. Expand $\sin x$ as a Maclaurin series:

$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$

$\frac◆LB◆x - \sin x◆RB◆◆LB◆x^3◆RB◆ = \frac◆LB◆x - \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right)◆RB◆◆LB◆x^3◆RB◆ = \frac◆LB◆\frac{x^3}{6} - \frac{x^5}{120} + \cdots◆RB◆◆LB◆x^3◆RB◆ = \frac{1}{6} - \frac{x^2}{120} + \cdots$

Taking $x \to 0$: $\boxed{\displaystyle\lim_{x \to 0} \frac◆LB◆x - \sin x◆RB◆◆LB◆x^3◆RB◆ = \frac{1}{6}}$

Example 8.7: Integration involving inverse trigonometric functions

Problem. Evaluate $\displaystyle\int \arcsin x\,dx$.

Solution. Use integration by parts with $u = \arcsin x$, $dv = dx$:

$du = \frac◆LB◆1◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆\,dx, \quad v = x$

$\int \arcsin x\,dx = x\arcsin x - \int \frac◆LB◆x◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆\,dx$

For the second integral, let $w = 1 - x^2$, $dw = -2x\,dx$:

$\int \frac◆LB◆x◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆\,dx = -\sqrt{1-x^2}$

$\boxed{\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C}$

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9. Common Pitfalls

| Pitfall | Correct Approach | | ------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------- | --------------------------------------------- | ----------------------------------------------- | | Forgetting the chain rule when differentiating composite inverse trig functions | Always write $\dfrac{d}{dx}\!\left[\arcsin(u)\right] = \dfrac◆LB◆u'◆RB◆◆LB◆\sqrt{1-u^2}◆RB◆$ | | Using $\ln | x |$ before checking if the integral is improper | Check for discontinuities in the interval first | | Forgetting $+C$ on every antiderivative | Every indefinite integral needs an arbitrary constant | | Applying reduction formulae without checking the base case | Always state $I_0$ or $I_1$ explicitly | | Confusing $\dfrac{d^n y}{dx^n}$ notation with $\left(\dfrac{dy}{dx}\right)^n$ | $\dfrac{d^n y}{dx^n}$ is the $n$-th derivative, not the $n$-th power |

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10. Additional Exam-Style Questions

Question 8

Using the substitution $u = e^x$, find $\displaystyle\int \frac{e^x}{e^{2x} + 1}\,dx$.

Solution

$u = e^x$, $du = e^x\,dx$.

$\int \frac{du}{u^2 + 1} = \arctan u + C = \boxed{\arctan(e^x) + C}$

Question 9

The reduction formula $I_n = \displaystyle\int_0^{\pi/4} \tan^n x\,dx$ satisfies $I_n = \dfrac{1}{n-1} - I_{n-2}$ for $n \geq 2$. Given $I_0 = \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$ and $I_1 = \dfrac{1}{2}\ln 2$, find $I_3$.

Solution

$I_3 = \dfrac{1}{2} - I_1 = \dfrac{1}{2} - \dfrac{1}{2}\ln 2 = \dfrac{1}{2}(1 - \ln 2)$.

To verify: $I_2 = \dfrac{1}{1} - I_0 = 1 - \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$. Then $I_3 = \dfrac{1}{2} - I_1 = \dfrac{1}{2} - \dfrac{1}{2}\ln 2$. Consistent. $\boxed{I_3 = \dfrac{1}{2}(1 - \ln 2)}$

Question 10

Find the area enclosed by the curve $x = t^2$, $y = t^3 - t$ for $-1 \leq t \leq 1$.

Solution

Using the parametric area formula $A = \displaystyle\int y\,\frac{dx}{dt}\,dt$:

$A = \int_{-1}^{1} (t^3 - t)(2t)\,dt = 2\int_{-1}^{1} (t^4 - t^2)\,dt = 2\left[\frac{t^5}{5} - \frac{t^3}{3}\right]_{-1}^{1}$

Since $t^5 - \frac{5}{3}t^3$ is odd (each term is odd), the integral from $-1$ to $1$ is zero.

$\boxed{A = 0}$ (the curve traces back over itself symmetrically).

Question 11

Prove that $\dfrac{d}{dx}\!\left[\arctan x\right] = \dfrac{1}{1+x^2}$ from first principles using implicit differentiation.

Solution

Let $y = \arctan x$, so $x = \tan y$. Differentiating implicitly with respect to $x$:

$1 = \sec^2 y \cdot \frac{dy}{dx}$

$\frac{dy}{dx} = \cos^2 y = \frac◆LB◆1◆RB◆◆LB◆\sec^2 y◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆1 + \tan^2 y◆RB◆ = \frac{1}{1 + x^2}$

$\blacksquare$

Question 12

Evaluate $\displaystyle\int_0^1 \frac◆LB◆\ln x◆RB◆◆LB◆1+x◆RB◆\,dx$, expressing your answer in terms of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$.

Solution

Expand $\dfrac{1}{1+x} = \displaystyle\sum_{n=0}^{\infty} (-1)^n x^n$ for $|x| < 1$:

$\int_0^1 \ln x \sum_{n=0}^{\infty} (-1)^n x^n\,dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln x\,dx$

Using integration by parts or the standard result $\displaystyle\int_0^1 x^n \ln x\,dx = -\frac{1}{(n+1)^2}$:

$= -\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2} = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$

This equals $-\dfrac◆LB◆\pi^2◆RB◆◆LB◆12◆RB◆$.

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11. Connections to Other Topics

11.1 Further calculus and differential equations

Integration techniques (substitution, parts, partial fractions) are essential tools for solving differential equations. See Differential Equations.

11.2 Calculus and Maclaurin series

Taylor and Maclaurin expansions provide powerful tools for evaluating integrals that cannot be found in closed form. See Maclaurin and Taylor Series.

11.3 Calculus and mechanics

Arc length and area calculations are used extensively in mechanics for work-energy problems. See Circular Motion.

11.4 Calculus and hyperbolic functions

The inverse hyperbolic functions arise naturally from integration: $\displaystyle\int \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2+a^2}◆RB◆ = \operatorname{arsinh}(x/a) + C$. See Hyperbolic Functions.

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12. Advanced Integration Techniques

12.1 Integration by parts — LIATE rule

When choosing $u$ and $dv$ for integration by parts, use the LIATE priority:

• Logarithmic functions
• Inverse trigonometric functions
• Algebraic functions (polynomials)
• Trigonometric functions
• Exponential functions

The function higher on the list should be chosen as $u$.

12.2 The Weierstrass substitution

For integrals involving rational functions of $\sin x$ and $\cos x$, the substitution $t = \tan(x/2)$ converts them to rational functions of $t$:

$\sin x = \dfrac{2t}{1+t^2}$, $\cos x = \dfrac{1-t^2}{1+t^2}$, $dx = \dfrac{2\,dt}{1+t^2}$.

12.3 Recognising standard integral forms

| Form | Result | | ------------------------------------------------------------- | ------------------------- | ---- | ---- | | $\displaystyle\int \frac{f'(x)}{f(x)}\,dx$ | $\ln | f(x) | + C$ | | $\displaystyle\int \frac◆LB◆f'(x)◆RB◆◆LB◆\sqrt{f(x)}◆RB◆\,dx$ | $2\sqrt{f(x)} + C$ | | $\displaystyle\int f(x) \cdot f'(x)\,dx$ | $\dfrac{[f(x)]^2}{2} + C$ |

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13. Summary of Key Results

Result	Formula
Integration by parts	$\displaystyle\int u\,dv = uv - \int v\,du$
Reduction formula (by parts)	Express $I_n$ in terms of $I_{n-1}$ or $I_{n-2}$
Arc length (Cartesian)	$s = \displaystyle\int_a^b \sqrt◆LB◆1+\left(\frac{dy}{dx}\right)^2◆RB◆\,dx$
Arc length (parametric)	$s = \displaystyle\int_\alpha^\beta \sqrt◆LB◆\dot{x}^2+\dot{y}^2◆RB◆\,dt$
Area under parametric curve	$A = \displaystyle\int y\frac{dx}{dt}\,dt$
Surface of revolution	$S = 2\pi\displaystyle\int_a^b y\sqrt{1+(y')^2}\,dx$
Derivative of $\arcsin x$	$\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{1-x^2}◆RB◆$
Derivative of $\arctan x$	$\dfrac{1}{1+x^2}$
Improper integral test	$\displaystyle\int_a^\infty f(x)\,dx = \lim_{b\to\infty}\int_a^b f(x)\,dx$

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14. Further Exam-Style Questions

Question 13

Using integration by parts, evaluate $\displaystyle\int x^3 e^{-x}\,dx$.

Solution

Let $u = x^3$, $dv = e^{-x}\,dx$. $du = 3x^2\,dx$, $v = -e^{-x}$.

$\int x^3 e^{-x}\,dx = -x^3 e^{-x} + 3\int x^2 e^{-x}\,dx$.

Repeating: $\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int xe^{-x}\,dx = -x^2 e^{-x} - 2xe^{-x} + 2\int e^{-x}\,dx$.

$= -x^2 e^{-x} - 2xe^{-x} - 2e^{-x}$.

Therefore: $\int x^3 e^{-x}\,dx = -x^3 e^{-x} - 3x^2 e^{-x} - 6xe^{-x} - 6e^{-x} + C$.

$\boxed{= -e^{-x}(x^3 + 3x^2 + 6x + 6) + C}$

Question 14

Find the arc length of the curve $y = \ln(\cos x)$ from $x = 0$ to $x = \pi/4$.

Solution

$y' = -\tan x$. $1 + (y')^2 = 1 + \tan^2 x = \sec^2 x$.

$s = \displaystyle\int_0^{\pi/4} \sec x\,dx = [\ln|\sec x + \tan x|]_0^{\pi/4} = \ln(\sqrt{2}+1) - \ln(1) = \boxed{\ln(\sqrt{2}+1)}$.

Question 15

Prove that $\displaystyle\int_0^{\pi/2} \sin^n x\,dx = \dfrac{n-1}{n} \cdot \dfrac{n-3}{n-2} \cdots \times \begin{cases} 1 & n \text{ odd} \\ \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ & n \text{ even}\end{cases}$ (Wallis' formula).

Solution

Let $I_n = \displaystyle\int_0^{\pi/2} \sin^n x\,dx$.

Integration by parts with $u = \sin^{n-1}x$, $dv = \sin x\,dx$:

$I_n = [-\cos x \sin^{n-1}x]_0^{\pi/2} + (n-1)\displaystyle\int_0^{\pi/2} \cos^2 x \sin^{n-2}x\,dx$

$= 0 + (n-1)\displaystyle\int_0^{\pi/2} (1-\sin^2 x)\sin^{n-2}x\,dx = (n-1)(I_{n-2} - I_n)$.

$nI_n = (n-1)I_{n-2}$, so $\boxed{I_n = \dfrac{n-1}{n}I_{n-2}}$.

Base cases: $I_0 = \pi/2$, $I_1 = 1$.

For even $n$: $I_n = \dfrac{n-1}{n} \cdot \dfrac{n-3}{n-2} \cdots \dfrac{1}{2} \cdot \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$.

For odd $n$: $I_n = \dfrac{n-1}{n} \cdot \dfrac{n-3}{n-2} \cdots \dfrac{2}{3} \cdot 1$. $\blacksquare$

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15. Advanced Topics

15.1 The gamma function and factorial

The gamma function extends the factorial: $\Gamma(n) = (n-1)!$ for positive integers, and $\Gamma(x) = \displaystyle\int_0^{\infty} t^{x-1}e^{-t}\,dt$ for $x > 0$.

Wallis' formula leads to the important result: $\Gamma(1/2) = \sqrt◆LB◆\pi◆RB◆$.

15.2 Frullani's integral

For suitable functions $f$: $\displaystyle\int_0^{\infty} \frac{f(ax)-f(bx)}{x}\,dx = (f(0)-f(\infty))\ln\frac{b}{a}$.

Example: $\displaystyle\int_0^{\infty} \frac{e^{-ax}-e^{-bx}}{x}\,dx = \ln\frac{b}{a}$.

15.3 Differentiation under the integral sign

Leibniz's rule: $\dfrac◆LB◆d◆RB◆◆LB◆d\alpha◆RB◆\displaystyle\int_a^b f(x,\alpha)\,dx = \int_a^b \frac◆LB◆\partial f◆RB◆◆LB◆\partial\alpha◆RB◆\,dx$.

This is a powerful technique for evaluating integrals that depend on a parameter.

15.4 Improper integrals — comparison test

If $0 \leq f(x) \leq g(x)$ for $x \geq a$ and $\displaystyle\int_a^{\infty} g(x)\,dx$ converges, then $\displaystyle\int_a^{\infty} f(x)\,dx$ also converges.

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16. Further Exam-Style Questions

Question 16

Evaluate $\displaystyle\int_0^{\infty} xe^{-x}\,dx$ and relate it to the mean of the exponential distribution.

Solution

Integration by parts with $u = x$, $dv = e^{-x}\,dx$:

$= [-xe^{-x}]_0^{\infty} + \displaystyle\int_0^{\infty} e^{-x}\,dx = 0 + 1 = \boxed{1}$.

This equals $E(X)$ for $X \sim \mathrm{Exp}(1)$, confirming the result $E(X) = 1/\lambda$ with $\lambda = 1$.

Question 17

Prove that $\displaystyle\int_0^{\pi/2} \sin^2 x\cos^2 x\,dx = \frac◆LB◆\pi◆RB◆◆LB◆16◆RB◆$.

Solution

$\sin^2 x\cos^2 x = \dfrac◆LB◆\sin^2 2x◆RB◆◆LB◆4◆RB◆ = \dfrac◆LB◆1-\cos 4x◆RB◆◆LB◆8◆RB◆$.

$\displaystyle\int_0^{\pi/2} \frac◆LB◆1-\cos 4x◆RB◆◆LB◆8◆RB◆\,dx = \frac{1}{8}\!\left[x-\frac◆LB◆\sin 4x◆RB◆◆LB◆4◆RB◆\right]_0^{\pi/2} = \frac{1}{8}\cdot\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ = \boxed{\dfrac◆LB◆\pi◆RB◆◆LB◆16◆RB◆}$. $\blacksquare$

Question 18

Use integration by parts twice to evaluate $\displaystyle\int e^x\cos x\,dx$.

Solution

$I = \displaystyle\int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - (-e^x\cos x + \int e^x\cos x\,dx)$.

$I = e^x\sin x + e^x\cos x - I$.

$2I = e^x(\sin x+\cos x)$.

$\boxed{I = \dfrac◆LB◆e^x(\sin x+\cos x)◆RB◆◆LB◆2◆RB◆ + C}$

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17. Further Exam-Style Questions

Question 19

Evaluate $\displaystyle\int_0^1 \frac{x^3}{1+x^2}\,dx$.

Solution

Let $u = 1+x^2$, $du = 2x\,dx$. Note $x^2 = u-1$, so $x^3\,dx = x^2 \cdot x\,dx = (u-1)\cdot\dfrac{du}{2}$.

$\displaystyle\int_0^1 \frac{x^3}{1+x^2}\,dx = \frac{1}{2}\int_1^2 \frac{u-1}{u}\,du = \frac{1}{2}\int_1^2 \!\left(1-\frac{1}{u}\right)du$

$= \frac{1}{2}\Big[u-\ln u\Big]_1^2 = \frac{1}{2}(2-\ln 2 - 1) = \boxed{\frac{1}{2}(1-\ln 2)}$.

Question 20

Prove that the function $F(x) = \displaystyle\int_0^x \frac{dt}{1+t^4}$ is increasing and bounded above.

Solution

$F'(x) = \dfrac{1}{1+x^4} > 0$ for all $x \geq 0$, so $F$ is strictly increasing. ✓

$F(x) < F(\infty) = \displaystyle\int_0^{\infty} \frac{dt}{1+t^4} < \int_0^{\infty} \frac{dt}{1+t^2} = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$. ✓

Therefore $F$ is increasing and bounded above by $\pi/2$. $\blacksquare$