Integration by Substitution (Edexcel A-Level Further Mathematics): Revision Notes

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5.2.5 Integration by Substitution

Introduction to Integration by Substitution

Integration by substitution is a powerful technique for solving integrals where a direct approach is challenging. It involves substituting part of the integral with a simpler variable to make the integral easier to solve. For functions involving expressions like a2−x2\sqrt{a^2 - x^2} or x2−a2\sqrt{x^2 - a^2}, trigonometric substitutions are particularly useful.

Trigonometric Substitutions for Integrals

For a2−x2\sqrt{a^2 - x^2}:

Use the substitution:

x=asin⁡θ,dx=acos⁡θ dθ,a2−x2=acos⁡θx = a \sin \theta, \quad dx = a \cos \theta \, d\theta, \quad \sqrt{a^2 - x^2} = a \cos \theta

This substitution is derived from the Pythagorean identity:

sin⁡2θ+cos⁡2θ=1 \sin^2 \theta + \cos^2 \theta = 1

For x2−a2\sqrt{x^2 - a^2}:

Use the substitution:

x=asec⁡θ,dx=asec⁡θtan⁡θ dθ,x2−a2=atan⁡θx = a \sec \theta, \quad dx = a \sec \theta \tan \theta \, d\theta, \quad \sqrt{x^2 - a^2} = a \tan \theta

This substitution is based on the identity:

sec⁡2θ−1=tan⁡2θ\sec^2 \theta - 1 = \tan^2 \theta

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

Use the substitution:

x=atan⁡θ,dx=asec⁡2θ dθ,x2+a2=asec⁡θx = a \tan \theta, \quad dx = a \sec^2 \theta \, d\theta, \quad \sqrt{x^2 + a^2} = a \sec \theta

This substitution uses:

1+tan⁡2θ=sec⁡2θ 1 + \tan^2 \theta = \sec^2 \theta

Worked Examples

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Example 1: Evaluate ∫4−x2 dx\int \sqrt{4 - x^2} \, dx

Step 1: Substitute x=2sin⁡θx = 2\sin \theta:

x=2sin⁡θ,dx=2cos⁡θ dθ,4−x2=2cos⁡θx = 2\sin \theta, \quad dx = 2\cos \theta \, d\theta, \quad \sqrt{4 - x^2} = 2\cos \theta

Step 2: Rewrite the integral:

∫4−x2 dx=∫2cos⁡θ⋅2cos⁡θ dθ=∫4cos⁡2θ dθ\int \sqrt{4 - x^2} \, dx = \int 2\cos \theta \cdot 2\cos \theta \, d\theta = \int 4\cos^2 \theta \, d\theta

Step 3: Simplify using a trigonometric identity:

Use cos⁡2θ=1+cos⁡2θ2\cos^2 \theta = \frac{1 + \cos 2\theta}{2}:

∫4cos⁡2θ dθ=∫2(1+cos⁡2θ) dθ=2∫1 dθ+2∫cos⁡2θ dθ\int 4\cos^2 \theta \, d\theta = \int 2(1 + \cos 2\theta) \, d\theta = 2\int 1 \, d\theta + 2\int \cos 2\theta \, d\theta

Step 4: Integrate term by term:

  • ∫1 dθ=θ\int 1 \, d\theta = \theta
  • ∫cos⁡2θ dθ=sin⁡2θ2\int \cos 2\theta \, d\theta = \frac{\sin 2\theta}{2} Combine:
∫4cos⁡2θ dθ=2θ+sin⁡2θ+C\int 4\cos^2 \theta \, d\theta = 2\theta + \sin 2\theta + C

Step 5: Back-substitute:

Using sin⁡θ=x2\sin \theta = \frac{x}{2}

and cos⁡θ=1−sin⁡2θ=1−x24=4−x22\cos \theta = \sqrt{1-\sin^2 \theta} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4-x^2}}{2}

  • θ=arcsin⁡(x2)\theta = \arcsin\left(\frac{x}{2}\right)
  • sin⁡2θ=2sin⁡θcos⁡θ=2(x2)(4−x22)=x4−x22.\sin 2\theta = 2\sin \theta \cos \theta = 2\left(\frac{x}{2}\right)\left(\frac{\sqrt{4-x^2}}{2}\right) = \frac{x\sqrt{4-x^2}}{2}. Substitute back:
∫4−x2 dx=2arcsin⁡(x2)+x4−x22+C\int \sqrt{4 - x^2} \, dx = 2\arcsin\left(\frac{x}{2}\right) + \frac{x\sqrt{4-x^2}}{2} + C
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Example 2: Evaluate ∫dxx2−9\int \frac{dx}{\sqrt{x^2 - 9}}

Step 1: Substitute x=3sec⁡θx = 3\sec \theta

x=3sec⁡θ,dx=3sec⁡θtan⁡θ dθ,x2−9=3tan⁡θx = 3\sec \theta, \quad dx = 3\sec \theta \tan \theta \, d\theta, \quad \sqrt{x^2 - 9} = 3\tan \theta

Step 2: Rewrite the integral:

∫dxx2−9=∫3sec⁡θtan⁡θ dθ3tan⁡θ\int \frac{dx}{\sqrt{x^2 - 9}} = \int \frac{3\sec \theta \tan \theta \, d\theta}{3\tan \theta}

Step 3: Simplify:

∫dxx2−9=∫sec⁡θ dθ\int \frac{dx}{\sqrt{x^2 - 9}} = \int \sec \theta \, d\theta

Step 4: Integrate sec⁡θsec⁡θsec⁡θ\sec \theta:

∫sec⁡θ dθ=ln⁡∣sec⁡θ+tan⁡θ∣+C\int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| + C

Step 5: Back-substitute:

Using x=3sec⁡θx = 3\sec \theta, we have:

  • sec⁡θ=x3\sec \theta = \frac{x}{3}
  • tan⁡θ=sec⁡2θ−1=(x3)2−1=x2−93\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\left(\frac{x}{3}\right)^2 - 1} = \frac{\sqrt{x^2 - 9}}{3} Substitute back:
∫dxx2−9=ln⁡∣x3+x2−93∣+C=ln⁡∣x+x2−93∣+C\int \frac{dx}{\sqrt{x^2 - 9}} = \ln\left|\frac{x}{3} + \frac{\sqrt{x^2 - 9}}{3}\right| + C = \ln\left|\frac{x + \sqrt{x^2 - 9}}{3}\right| + C
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Example 3: Evaluate ∫dxx2+4\int \frac{dx}{x^2 + 4}

Step 1: Substitute x=2tan⁡θx = 2\tan \theta

x=2tan⁡θ,dx=2sec⁡2θ dθ,x2+4=4sec⁡2θx = 2\tan \theta, \quad dx = 2\sec^2 \theta \, d\theta, \quad x^2 + 4 = 4\sec^2 \theta

Step 2: Rewrite the integral:

∫dxx2+4=∫2sec⁡2θ dθ4sec⁡2θ\int \frac{dx}{x^2 + 4} = \int \frac{2\sec^2 \theta \, d\theta}{4\sec^2 \theta}

Step 3: Simplify:

∫dxx2+4=12∫dθ\int \frac{dx}{x^2 + 4} = \frac{1}{2} \int d\theta

Step 4: Integrate:

∫dxx2+4=θ2+C\int \frac{dx}{x^2 + 4} = \frac{\theta}{2} + C

Step 5: Back-substitute:

Using x=2tan⁡θx = 2\tan \theta, we have θ=arctan⁡(x2)\theta = \arctan\left(\frac{x}{2}\right):

∫dxx2+4=12arctan⁡(x2)+C\int \frac{dx}{x^2 + 4} = \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C

Note Summary

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Common Mistakes:

  1. Forgetting the substitution derivatives: Ensure dxdx is replaced correctly.

  2. Incorrect trigonometric identities: Verify sin⁡2θ+cos⁡2θ=1,sec⁡2θ−1=tan⁡2θ\sin^2 \theta + \cos^2 \theta = 1, \sec^2 \theta - 1 = \tan^2 \theta, etc.

  3. Not simplifying fully after back-substitution: Return to the original variable xx wherever possible.

  4. Mismanaging bounds in definite integrals: Adjust integration limits to match the substitution.

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Key Formulas:

  1. Trigonometric Substitutions:
  • a2−x2:x=asin⁡θ,dx=acos⁡θ dθ\sqrt{a^2 - x^2}: x = a\sin \theta, dx = a\cos \theta \, d\theta
  • x2−a2:x=asec⁡θ,dx=asec⁡θtan⁡θ dθ\sqrt{x^2 - a^2}: x = a\sec \theta, dx = a\sec \theta \tan \theta \, d\theta
  • x2+a2:x=atan⁡θ,dx=asec⁡2θ dθ\sqrt{x^2 + a^2}: x = a\tan \theta, dx = a\sec^2 \theta \, d\theta
  1. Standard Integrals:
  • ∫dxa2−x2=arcsin⁡(xa)+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C
  • ∫dxx2−a2=ln⁡∣x+x2−a2∣+C\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C
  • ∫dxx2+a2=1aarctan⁡(xa)+C\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C

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