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Integration Techniques

Substitution Method for Integration

Master u-substitution and trigonometric substitution for complex integrals

1. U-Substitution (Chain Rule Reversal)

Theorem 1.1: Substitution Rule

If u=g(x)u = g(x) is a differentiable function, then:

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

This is the reverse of the chain rule for differentiation.

Example 1.1: Basic U-Substitution

Find: ∫2xcos⁡(x2) dx\int 2x\cos(x^2)\,dx

Solution:

Step 1: Let u=x2u = x^2, then du=2x dxdu = 2x\,dx

Step 2: Substitute:

∫2xcos⁡(x2) dx=∫cos⁡(u) du=sin⁡(u)+C\int 2x\cos(x^2)\,dx = \int \cos(u)\,du = \sin(u) + C

Step 3: Back-substitute:

=sin⁡(x2)+C= \sin(x^2) + C
Example 1.2: Adjusting Constants

Find: ∫x1−x2 dx\int x\sqrt{1-x^2}\,dx

Solution:

Step 1: Let u=1−x2u = 1-x^2, then du=−2x dxdu = -2x\,dx

Step 2: We have x dxx\,dx but need −2x dx-2x\,dx. Multiply by −1/2-1/2:

∫x1−x2 dx=−12∫u du\int x\sqrt{1-x^2}\,dx = -\frac{1}{2}\int \sqrt{u}\,du

Step 3: Integrate:

=−12⋅23u3/2+C=−13(1−x2)3/2+C= -\frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C = -\frac{1}{3}(1-x^2)^{3/2} + C

2. Trigonometric Substitution

Standard Trigonometric Substitutions

For a2−x2\sqrt{a^2 - x^2}:

Let x=asin⁡θx = a\sin\theta, then a2−x2=acos⁡θ\sqrt{a^2-x^2} = a\cos\theta

For a2+x2\sqrt{a^2 + x^2}:

Let x=atan⁡θx = a\tan\theta, then a2+x2=asec⁡θ\sqrt{a^2+x^2} = a\sec\theta

For x2−a2\sqrt{x^2 - a^2}:

Let x=asec⁡θx = a\sec\theta, then x2−a2=atan⁡θ\sqrt{x^2-a^2} = a\tan\theta

Example 2.1: Trigonometric Substitution

Find: ∫11−x2 dx\int \frac{1}{\sqrt{1-x^2}}\,dx

Solution:

Step 1: Let x=sin⁡θx = \sin\theta, then dx=cos⁡θ dθdx = \cos\theta\,d\theta

Step 2: Substitute:

1−x2=1−sin⁡2θ=cos⁡θ\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \cos\theta
∫11−x2 dx=∫cos⁡θcos⁡θ dθ=∫dθ=θ+C\int \frac{1}{\sqrt{1-x^2}}\,dx = \int \frac{\cos\theta}{\cos\theta}\,d\theta = \int d\theta = \theta + C

Step 3: Back-substitute: θ=arcsin⁡x\theta = \arcsin x

=arcsin⁡x+C= \arcsin x + C
Example 2.2: With Quadratic

Find: ∫1x2x2−4 dx\int \frac{1}{x^2\sqrt{x^2-4}}\,dx

Solution:

Step 1: Let x=2sec⁡θx = 2\sec\theta, then dx=2sec⁡θtan⁡θ dθdx = 2\sec\theta\tan\theta\,d\theta

Step 2: Substitute:

x2−4=4sec⁡2θ−4=2tan⁡θ\sqrt{x^2-4} = \sqrt{4\sec^2\theta-4} = 2\tan\theta
∫1x2x2−4 dx=∫2sec⁡θtan⁡θ4sec⁡2θ⋅2tan⁡θ dθ\int \frac{1}{x^2\sqrt{x^2-4}}\,dx = \int \frac{2\sec\theta\tan\theta}{4\sec^2\theta \cdot 2\tan\theta}\,d\theta
=14∫cos⁡θ dθ=14sin⁡θ+C= \frac{1}{4}\int \cos\theta\,d\theta = \frac{1}{4}\sin\theta + C

Step 3: Back-substitute using sin⁡θ=x2−4x\sin\theta = \frac{\sqrt{x^2-4}}{x}:

=x2−44x+C= \frac{\sqrt{x^2-4}}{4x} + C