Integration Techniques

Antidifferentiation Review

Integration (antidifferentiation) is the reverse process of differentiation. If $F'(x) = f(x)$, then:

$$\int f(x)\,dx = F(x) + C$$

where $C$ is the constant of integration.

Basic Integration Rules

Rule	Formula
Power rule	$\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
Reciprocal	$\displaystyle\int \frac{1}{x}\,dx = \ln\|x\| + C$
Constant multiple	$\displaystyle\int kf(x)\,dx = k\int f(x)\,dx$
Sum/difference	$\displaystyle\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$

Standard Integrals

$f(x)$	$\int f(x)\,dx$
$e^x$	$e^x + C$
$a^x$	$\dfrac{a^x}{\ln a} + C$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$
$\csc^2 x$	$-\cot x + C$
$\sec x \tan x$	$\sec x + C$
$\dfrac{1}{\sqrt{1-x^2}}$	$\arcsin x + C$
$\dfrac{1}{1+x^2}$	$\arctan x + C$

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The Fundamental Theorem of Calculus

First Fundamental Theorem

If $f$ is continuous on $[a, b]$ and $F'(x) = f(x)$, then:

$$\int_a^b f(x)\,dx = F(b) - F(a)$$

Second Fundamental Theorem

If $f$ is continuous on an interval containing $a$, then:

\frac{d}`\{dx}`\left[\int_a^x f(t)\,dt\right] = f(x)

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Definite Integrals

Properties

For continuous functions $f$ and $g$ on $[a, b]$:

1. $\displaystyle\int_a^b kf(x)\,dx = k\int_a^b f(x)\,dx$
2. $\displaystyle\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$
3. $\displaystyle\int_a^a f(x)\,dx = 0$
4. $\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$
5. $\displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ for any $c$

Even and Odd Functions

If $f$ is even ($f(-x) = f(x)$):

$$\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$$

If $f$ is odd ($f(-x) = -f(x)$):

$$\int_{-a}^{a} f(x)\,dx = 0$$

Exam Tip

Always check if a function is odd or even before integrating over a symmetric interval. This can save significant computation time.

Example

Evaluate $\displaystyle\int_{-2}^{2} x^3\cos(x^2)\,dx$.

Since $x^3\cos(x^2)$ is odd (odd $\times$ even $=$ odd):

$$\int_{-2}^{2} x^3\cos(x^2)\,dx = 0$$

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Integration by Substitution

Method

When the integrand contains a function and its derivative (or a composite function), substitute $u = g(x)$ where $g(x)$ is the "inner" function:

$$\int f(g(x))g'(x)\,dx = \int f(u)\,du$$

Steps

1. Choose $u = g(x)$ (the inner function).
2. Compute $\dfrac{du}{dx} = g'(x)$, so $du = g'(x)\,dx$.
3. Rewrite the integral in terms of $u$.
4. Integrate with respect to $u$.
5. Substitute back $u = g(x)$.

Example

Evaluate $\displaystyle\int 2x\sqrt{x^2+1}\,dx$.

Let $u = x^2 + 1$, so $\dfrac{du}{dx} = 2x$, giving $du = 2x\,dx$.

$$\int \sqrt{u}\,du = \int u^{1/2}\,du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x^2+1)^{3/2} + C$$

Example

Evaluate $\displaystyle\int_0^1 xe^{x^2}\,dx$.

Let $u = x^2$, so $du = 2x\,dx$.

When $x = 0$: $u = 0$. When $x = 1$: $u = 1$.

$$\int_0^1 xe^{x^2}\,dx = \frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}\big[e^u\big]_0^1 = \frac{1}{2}(e - 1)$$

Trigonometric Substitutions

Certain integrals can be simplified using trigonometric substitutions:

Expression	Substitution	Identity
$\sqrt{a^2 - x^2}$	$x = a\sin\theta$	$1 - \sin^2\theta = \cos^2\theta$
$\sqrt{a^2 + x^2}$	$x = a\tan\theta$	$1 + \tan^2\theta = \sec^2\theta$
$\sqrt{x^2 - a^2}$	$x = a\sec\theta$	$\sec^2\theta - 1 = \tan^2\theta$

Example

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

This is a standard integral: $\arcsin\!\left(\dfrac{x}{2}\right) + C$.

Alternatively, let $x = 2\sin\theta$, so $dx = 2\cos\theta\,d\theta$:

$$\int \frac{2\cos\theta}{2\cos\theta}\,d\theta = \int 1\,d\theta = \theta + C = \arcsin\!\left(\frac{x}{2}\right) + C$$

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Integration by Parts

For the product of two functions:

$$\int u\,dv = uv - \int v\,du$$

Choosing $u$ and $dv$

Use the LIATE rule (prioritise choosing $u$ from):

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

Example

Evaluate $\displaystyle\int x e^x\,dx$.

Let $u = x$ (algebraic) and $dv = e^x\,dx$.

Then $du = dx$ and $v = e^x$.

$$\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x - 1) + C$$

Example

Evaluate $\displaystyle\int x\sin x\,dx$.

Let $u = x$ and $dv = \sin x\,dx$.

Then $du = dx$ and $v = -\cos x$.

$$\int x\sin x\,dx = -x\cos x - \int (-\cos x)\,dx = -x\cos x + \sin x + C$$

Repeated Integration by Parts

For integrals like $\displaystyle\int x^2 e^x\,dx$ or $\displaystyle\int x^2 \sin x\,dx$, apply integration by parts repeatedly.

Example

Evaluate $\displaystyle\int x^2 e^x\,dx$.

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

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

Second application on $\int 2x e^x\,dx$: $u = 2x$, $dv = e^x\,dx$:

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

Combining:

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

Cyclic Integration by Parts

Integrals like $\displaystyle\int e^x \sin x\,dx$ require applying integration by parts twice and solving algebraically.

Example

Evaluate $\displaystyle\int e^x \sin x\,dx$.

Let $u = e^x$, $dv = \sin x\,dx$. Then $du = e^x\,dx$, $v = -\cos x$.

$$I = -e^x\cos x + \int e^x\cos x\,dx$$

Now let $u = e^x$, $dv = \cos x\,dx$. Then $du = e^x\,dx$, $v = \sin x$.

$$\int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - I$$

Substituting back:

$$I = -e^x\cos x + e^x\sin x - I$$$$2I = e^x(\sin x - \cos x)$$$$I = \frac{e^x(\sin x - \cos x)}{2} + C$$

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Partial Fractions

When the integrand is a rational function with a denominator that can be factorised, decompose into partial fractions.

Types of Decomposition

Denominator	Partial Fractions
$(ax+b)(cx+d)$	$\dfrac{A}{ax+b} + \dfrac{B}{cx+d}$
$(ax+b)^2$	$\dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2}$
$(ax+b)(x^2+c)$	$\dfrac{A}{ax+b} + \dfrac{Bx+C}{x^2+c}$

Example

Evaluate $\displaystyle\int \frac{5x+1}{(x+1)(x-2)}\,dx$.

$$\frac{5x+1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$$$$5x + 1 = A(x-2) + B(x+1)$$

$x = -1$: $-5 + 1 = A(-3) \implies A = \dfrac{4}{3}$.

$x = 2$: $10 + 1 = B(3) \implies B = \dfrac{11}{3}$.

$$\int \frac{5x+1}{(x+1)(x-2)}\,dx = \frac{4}{3}\ln\|x+1\| + \frac{11}{3}\ln\|x-2\| + C$$

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Area Under Curves

Area Between a Curve and the $x$-axis

The signed area between $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$ is:

$$\mathrm{Area} = \int_a^b f(x)\,dx$$

If the curve crosses the $x$-axis, split the integral at the zeros and take the absolute value of each part.

Area Between Two Curves

The area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ (where $f(x) \ge g(x)$):

$$\mathrm{Area} = \int_a^b [f(x) - g(x)]\,dx$$

Example

Find the area enclosed by $y = x^2$ and $y = 2x$.

Intersection points: $x^2 = 2x \implies x(x-2) = 0 \implies x = 0$ or $x = 2$.

For $0 \le x \le 2$: $2x \ge x^2$.

$$\mathrm{Area} = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$$

Example

Find the area enclosed by $y = x^2 - 4$ and the $x$-axis.

Zeros: $x^2 - 4 = 0 \implies x = \pm 2$.

$$\mathrm{Area} = \int_{-2}^{2} |x^2 - 4|\,dx = 2\int_0^2 (4 - x^2)\,dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right) = \frac{32}{3}$$

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Volume of Revolution

Rotation about the $x$-axis

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

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

Rotation about the $y$-axis

The volume generated by rotating $x = g(y)$ from $y = c$ to $y = d$ about the $y$-axis:

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

Alternatively, using the shell method about the $y$-axis:

$$V = 2\pi \int_a^b x |f(x)|\,dx$$

Example

Find the volume generated by rotating the region bounded by $y = \sqrt{x}$, $x = 4$, and the $x$-axis 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

Find the volume generated by rotating the region bounded by $y = x^2$, $y = 0$, and $x = 1$ about the $y$-axis.

Using the disc method (in terms of $y$):

$x = \sqrt{y}$, from $y = 0$ to $y = 1$.

$$V = \pi\int_0^1 (\sqrt{y})^2\,dy = \pi\int_0^1 y\,dy = \pi\left[\frac{y^2}{2}\right]_0^1 = \frac{\pi}{2}$$

Volume between two curves

When rotating the region between $y = f(x)$ and $y = g(x)$ about the $x$-axis:

$$V = \pi\int_a^b \left([f(x)]^2 - [g(x)]^2\right)\,dx$$

Example

Find the volume generated by rotating the region between $y = x$ and $y = x^2$ about the $x$-axis.

Intersection: $x = x^2 \implies x = 0$ or $x = 1$.

$$V = \pi\int_0^1 \left(x^2 - x^4\right)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$$

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Kinematics Applications

Displacement, Velocity, and Acceleration

• Velocity is the derivative of displacement: $v = \dfrac{ds}{dt}$
• Acceleration is the derivative of velocity: $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$
• Displacement from velocity: $s = \displaystyle\int v\,dt$
• Velocity from acceleration: $v = \displaystyle\int a\,dt$

Total Distance vs Displacement

$$\mathrm{Displacement} = \int_{t_1}^{t_2} v(t)\,dt$$ $$\mathrm{Total distance} = \int_{t_1}^{t_2} |v(t)|\,dt$$

Example

A particle moves with velocity $v(t) = t^2 - 4t + 3\mathrm{ m/s}$ for $0 \le t \le 5$.

(a) Find the displacement.

$$s = \int_0^5 (t^2 - 4t + 3)\,dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^5 = \frac{125}{3} - 50 + 15 = \frac{40}{3}\mathrm{ m}$$

(b) Find the total distance travelled.

Find when $v = 0$: $t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0 \implies t = 1, 3$.

Interval	Sign of $v$	Motion
$0 \lt t \lt 1$	$+$	Forward
$1 \lt t \lt 3$	$-$	Backward
$3 \lt t \lt 5$	$+$	Forward

$$\mathrm{Distance} = \int_0^1 v\,dt + \left|\int_1^3 v\,dt\right| + \int_3^5 v\,dt$$$$= \frac{4}{3} + \left|-\frac{4}{3}\right| + \frac{20}{3} = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}\mathrm{ m}$$

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Improper Integrals

Type 1: Infinite Limits

$$\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$$

Type 2: Discontinuous Integrand

$$\int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx$$

An improper integral converges if the limit exists (is finite) and diverges otherwise.

Example

Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx$ converges.

$$\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b \to \infty}\int_1^b x^{-2}\,dx = \lim_{b \to \infty}\left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty}\left(-\frac{1}{b} + 1\right) = 1$$

The integral converges to $1$.

Example

Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x}\,dx$ converges.

$$\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty}[\ln x]_1^b = \lim_{b \to \infty}\ln b = \infty$$

The integral diverges.

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IB Exam-Style Questions

Question 1 (Paper 1 style)

Evaluate $\displaystyle\int_0^{\pi/2} \sin^2 x\,dx$.

Using the identity $\sin^2 x = \dfrac{1 - \cos 2x}{2}$:

$$\int_0^{\pi/2} \frac{1 - \cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{1}{2}\left(\frac{\pi}{2} - 0\right) = \frac{\pi}{4}$$

Question 2 (Paper 2 style)

Let $R$ be the region bounded by the curve $y = x(x-2)$ and the $x$-axis.

(a) Find the area of $R$.

Zeros: $x = 0$ and $x = 2$. The curve is below the axis (opens upward).

$$\mathrm{Area} = \int_0^2 |x(x-2)|\,dx = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$$

(b) Find the volume when $R$ is rotated $360\degree$ about the $x$-axis.

$$V = \pi\int_0^2 [x(x-2)]^2\,dx = \pi\int_0^2 x^2(x-2)^2\,dx$$ $$= \pi\int_0^2 (x^4 - 4x^3 + 4x^2)\,dx = \pi\left[\frac{x^5}{5} - x^4 + \frac{4x^3}{3}\right]_0^2$$ $$= \pi\left(\frac{32}{5} - 16 + \frac{32}{3}\right) = \pi \cdot \frac{16}{15} = \frac{16\pi}{15}$$

Question 3 (Paper 2 style)

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

Factorise the denominator: $x^2 + 3x + 2 = (x+1)(x+2)$.

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

$x = -1$: $1 = A \implies A = 1$.

$x = -2$: $-1 = -B \implies B = 1$.

$$\int \frac{2x+3}{x^2+3x+2}\,dx = \ln\|x+1\| + \ln\|x+2\| + C = \ln\|(x+1)(x+2)\| + C$$

Question 4 (Paper 1 style)

Use integration by parts to evaluate $\displaystyle\int_0^1 xe^{2x}\,dx$.

Let $u = x$, $dv = e^{2x}\,dx$. Then $du = dx$, $v = \dfrac{1}{2}e^{2x}$.

$$\int_0^1 xe^{2x}\,dx = \left[\frac{1}{2}xe^{2x}\right]_0^1 - \frac{1}{2}\int_0^1 e^{2x}\,dx = \frac{e^2}{2} - \frac{1}{2}\left[\frac{e^{2x}}{2}\right]_0^1$$ $$= \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}$$

Question 5 (Paper 2 style)

A particle moves in a straight line with acceleration $a(t) = 6t - 2\mathrm{ m/s}^2$. At $t = 0$, the velocity is $4\mathrm{ m/s}$ and the displacement is $0\mathrm{ m}$.

(a) Find the velocity function.

$$v(t) = \int (6t-2)\,dt = 3t^2 - 2t + C$$

$v(0) = 4 \implies C = 4$.

$$v(t) = 3t^2 - 2t + 4$$

(b) Find the displacement function.

$$s(t) = \int (3t^2 - 2t + 4)\,dt = t^3 - t^2 + 4t + D$$

$s(0) = 0 \implies D = 0$.

$$s(t) = t^3 - t^2 + 4t$$

(c) Find the total distance travelled in the first 3 seconds.

Check if $v = 0$: $3t^2 - 2t + 4 = 0$. Discriminant $= 4 - 48 \lt 0$, so $v \gt 0$ always.

$$\mathrm{Distance} = \int_0^3 v\,dt = \int_0^3 (3t^2 - 2t + 4)\,dt = \left[t^3 - t^2 + 4t\right]_0^3 = 27 - 9 + 12 = 30\mathrm{ m}$$

Question 6 (Paper 2 style)

The region bounded by $y = e^x$, $y = 1$, $x = 0$, and $x = 2$ is rotated about the $x$-axis. Find the volume.

$$V = \pi\int_0^2 [(e^x)^2 - 1^2]\,dx = \pi\int_0^2 (e^{2x} - 1)\,dx = \pi\left[\frac{e^{2x}}{2} - x\right]_0^2$$ $$= \pi\left(\frac{e^4}{2} - 2 - \frac{1}{2} + 0\right) = \pi\left(\frac{e^4}{2} - \frac{5}{2}\right) = \frac{\pi(e^4 - 5)}{2}$$

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Summary Table of Techniques

Technique	When to Use	Key Idea
Direct	Simple power functions, standard forms	Apply basic rules directly
Substitution	Composite functions, $f(g(x))g'(x)$	Let $u = g(x)$
By parts	Product of different function types	$\int u\,dv = uv - \int v\,du$
Partial fractions	Rational functions, factorisable denominator	Decompose then integrate each term
Trig substitution	$\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$	Replace with trig function

Exam Strategy

When facing an integral, first check if it is a standard form. If not, consider substitution (especially if you see a function and its derivative). If it is a product of different function types, use integration by parts. If it is a rational function, consider partial fractions.

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Additional Integration Techniques

Integrals Involving Trigonometric Identities

Many trigonometric integrals require using identities to simplify before integrating.

Powers of Sine and Cosine

Odd power of sine: Factor out one $\sin x$, convert the rest to cosines using $\sin^2 x = 1 - \cos^2 x$, then substitute $u = \cos x$.

Example

Evaluate $\displaystyle\int \sin^3 x\,dx$.

$$\int \sin^2 x \sin x\,dx = \int (1 - \cos^2 x)\sin x\,dx$$

Let $u = \cos x$, $du = -\sin x\,dx$:

$$= -\int (1 - u^2)\,du = -u + \frac{u^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C$$

Even powers of sine or cosine: Use the half-angle formulas.

$$\sin^2 x = \frac{1 - \cos 2x}{2}, \quad \cos^2 x = \frac{1 + \cos 2x}{2}$$

Example

Evaluate $\displaystyle\int \cos^4 x\,dx$.

$$\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}$$$$= \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8}$$$$\int \cos^4 x\,dx = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$$

Products of Sine and Cosine

For $\displaystyle\int \sin mx \cos nx\,dx$, use the product-to-sum identities:

$$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$$ $$\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$$ $$\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$$

Example

Evaluate $\displaystyle\int \sin 3x \cos 5x\,dx$.

$$\sin 3x \cos 5x = \frac{1}{2}[\sin 8x + \sin(-2x)] = \frac{1}{2}(\sin 8x - \sin 2x)$$$$\int \sin 3x \cos 5x\,dx = \frac{1}{2}\left(-\frac{\cos 8x}{8} + \frac{\cos 2x}{2}\right) + C$$

Integrals of the Form $\displaystyle\int \frac{1}{a^2 + x^2}\,dx$ and Related

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

Example

Evaluate $\displaystyle\int \frac{1}{4 + 9x^2}\,dx$.

Rewrite as: $\displaystyle\int \frac{1}{4 + (3x)^2}\,dx$.

Let $u = 3x$, $du = 3\,dx$:

$$= \frac{1}{3}\int \frac{1}{4 + u^2}\,du = \frac{1}{3} \cdot \frac{1}{2}\arctan\frac{u}{2} + C = \frac{1}{6}\arctan\frac{3x}{2} + C$$

Integrals Involving $e^x$ and $\ln x$

Integrals with $e^x$ and Polynomials

Use integration by parts when $e^x$ is multiplied by a polynomial.

Integrals with $\ln x$

Use integration by parts with $u = \ln x$ and $dv = dx$:

$$\int \ln x\,dx = x\ln x - x + C$$

Example

Evaluate $\displaystyle\int x^2 \ln x\,dx$.

Let $u = \ln x$, $dv = x^2\,dx$. Then $du = \dfrac{1}{x}\,dx$, $v = \dfrac{x^3}{3}$.

$$\int x^2 \ln x\,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx$$$$= \frac{x^3}{3}\ln x - \frac{x^3}{9} + C$$

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Arc Length

The arc length of a curve $y = f(x)$ from $x = a$ to $x = b$:

L = \int_a^b \sqrt{1 + \left(\frac`\{dy}``\{dx}`\right)^2}\,dx

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

L = \int_{t_1}^{t_2} \sqrt{\left(\frac`\{dx}``\{dt}`\right)^2 + \left(\frac`\{dy}``\{dt}`\right)^2}\,dt

Example

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

\frac`\{dy}``\{dx}` = \frac{x^2}{2} - \frac{1}{2x^2}\left(\frac`\{dy}``\{dx}`\right)^2 = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}1 + \left(\frac`\{dy}``\{dx}`\right)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2$$L = \int_1^3 \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)\,dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^3$$$$= \left(\frac{27}{6} - \frac{1}{6}\right) - \left(\frac{1}{6} - \frac{1}{2}\right) = \frac{26}{6} + \frac{1}{3} = \frac{14}{3}$$

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Mean Value of a Function

The mean value of a function $f(x)$ over $[a, b]$:

$$\bar{f} = \frac{1}{b - a}\int_a^b f(x)\,dx$$

Example

Find the mean value of $f(x) = x^2$ over $[0, 3]$.

$$\bar{f} = \frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \times 9 = 3$$

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Additional Exam-Style Questions

Question 7 (Paper 2 style)

Evaluate $\displaystyle\int_0^{\pi/4} x\cos 2x\,dx$.

Let $u = x$, $dv = \cos 2x\,dx$. Then $du = dx$, $v = \dfrac{\sin 2x}{2}$.

$$\int_0^{\pi/4} x\cos 2x\,dx = \left[\frac{x\sin 2x}{2}\right]_0^{\pi/4} - \frac{1}{2}\int_0^{\pi/4}\sin 2x\,dx$$ $$= \left(\frac{\pi/4 \times 1}{2} - 0\right) - \frac{1}{2}\left[-\frac{\cos 2x}{2}\right]_0^{\pi/4}$$ $$= \frac{\pi}{8} - \frac{1}{2}\left(-\frac{0}{2} + \frac{1}{2}\right) = \frac{\pi}{8} - \frac{1}{4}$$

Question 8 (Paper 2 style)

Find the area enclosed by the curves $y = x^3$ and $y = \sqrt{x}$.

Intersection: $x^3 = \sqrt{x} \implies x^6 = x \implies x(x^5 - 1) = 0 \implies x = 0$ or $x = 1$.

For $0 \le x \le 1$: $\sqrt{x} \ge x^3$.

$$\mathrm{Area} = \int_0^1 (\sqrt{x} - x^3)\,dx = \left[\frac{2x^{3/2}}{3} - \frac{x^4}{4}\right]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$$

Question 9 (Paper 1 style)

Evaluate $\displaystyle\int \frac{x}{x^2 + 2x + 5}\,dx$.

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

Let $u = x^2 + 2x + 5$, $du = (2x + 2)\,dx$.

We need to split: $\dfrac{x}{x^2+2x+5} = \dfrac{(2x+2)/2 - 1}{x^2+2x+5} = \dfrac{1}{2} \cdot \dfrac{2x+2}{x^2+2x+5} - \dfrac{1}{x^2+2x+5}$.

$$\int \frac{x}{x^2+2x+5}\,dx = \frac{1}{2}\ln(x^2+2x+5) - \int \frac{1}{(x+1)^2 + 4}\,dx$$ $$= \frac{1}{2}\ln(x^2+2x+5) - \frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C$$

Question 10 (Paper 2 style)

The region bounded by $y = \ln x$, $y = 0$, and $x = e$ is rotated $360\degree$ about the $x$-axis. Find the volume.

$$V = \pi\int_1^e (\ln x)^2\,dx$$

Using integration by parts with $u = (\ln x)^2$, $dv = dx$:

$$= \pi\left[x(\ln x)^2\right]_1^e - 2\pi\int_1^e \ln x\,dx$$ $$= \pi(e - 0) - 2\pi[x\ln x - x]_1^e = \pi e - 2\pi(e - e + 1) = \pi e - 2\pi = \pi(e - 2)$$

For the A-Level treatment of this topic, see Integration.

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