Photo AI

Last Updated Sep 27, 2025

Integration by Substitution Simplified Revision Notes

Revision notes with simplified explanations to understand Integration by Substitution quickly and effectively.

user avatar
user avatar
user avatar
user avatar
user avatar

266+ students studying

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 a2x2\sqrt{a^2 - x^2} or x2a2\sqrt{x^2 - a^2}, trigonometric substitutions are particularly useful.

Trigonometric Substitutions for Integrals

For a2x2\sqrt{a^2 - x^2}:

Use the substitution:

x=asinθ,dx=acosθdθ,a2x2=acosθx = 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:

sin2θ+cos2θ=1 \sin^2 \theta + \cos^2 \theta = 1

For x2a2\sqrt{x^2 - a^2}:

Use the substitution:

x=asecθ,dx=asecθtanθdθ,x2a2=atanθx = 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:

sec2θ1=tan2θ\sec^2 \theta - 1 = \tan^2 \theta

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

Use the substitution:

x=atanθ,dx=asec2θdθ,x2+a2=asecθx = a \tan \theta, \quad dx = a \sec^2 \theta \, d\theta, \quad \sqrt{x^2 + a^2} = a \sec \theta

This substitution uses:

1+tan2θ=sec2θ 1 + \tan^2 \theta = \sec^2 \theta

Worked Examples

lightbulbExample

Example 1: Evaluate 4x2dx\int \sqrt{4 - x^2} \, dx


Step 1: Substitute x=2sinθx = 2\sin \theta:

x=2sinθ,dx=2cosθdθ,4x2=2cosθx = 2\sin \theta, \quad dx = 2\cos \theta \, d\theta, \quad \sqrt{4 - x^2} = 2\cos \theta

Step 2: Rewrite the integral:

4x2dx=2cosθ2cosθdθ=4cos2θdθ\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 cos2θ=1+cos2θ2\cos^2 \theta = \frac{1 + \cos 2\theta}{2}:

4cos2θdθ=2(1+cos2θ)dθ=21dθ+2cos2θdθ\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:

  • 1dθ=θ\int 1 \, d\theta = \theta
  • cos2θdθ=sin2θ2\int \cos 2\theta \, d\theta = \frac{\sin 2\theta}{2} Combine:
4cos2θdθ=2θ+sin2θ+C\int 4\cos^2 \theta \, d\theta = 2\theta + \sin 2\theta + C

Step 5: Back-substitute:

Using sinθ=x2\sin \theta = \frac{x}{2}

and cosθ=1sin2θ=1x24=4x22\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)
  • sin2θ=2sinθcosθ=2(x2)(4x22)=x4x22.\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:
4x2dx=2arcsin(x2)+x4x22+C\int \sqrt{4 - x^2} \, dx = 2\arcsin\left(\frac{x}{2}\right) + \frac{x\sqrt{4-x^2}}{2} + C
lightbulbExample

Example 2: Evaluate dxx29\int \frac{dx}{\sqrt{x^2 - 9}}


Step 1: Substitute x=3secθx = 3\sec \theta

x=3secθ,dx=3secθtanθdθ,x29=3tanθx = 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:

dxx29=3secθtanθdθ3tanθ\int \frac{dx}{\sqrt{x^2 - 9}} = \int \frac{3\sec \theta \tan \theta \, d\theta}{3\tan \theta}

Step 3: Simplify:

dxx29=secθdθ\int \frac{dx}{\sqrt{x^2 - 9}} = \int \sec \theta \, d\theta

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

secθdθ=lnsecθ+tanθ+C\int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| + C

Step 5: Back-substitute:

Using x=3secθx = 3\sec \theta, we have:

  • secθ=x3\sec \theta = \frac{x}{3}
  • tanθ=sec2θ1=(x3)21=x293\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\left(\frac{x}{3}\right)^2 - 1} = \frac{\sqrt{x^2 - 9}}{3} Substitute back:
dxx29=lnx3+x293+C=lnx+x293+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
lightbulbExample

Example 3: Evaluate dxx2+4\int \frac{dx}{x^2 + 4}


Step 1: Substitute x=2tanθx = 2\tan \theta

x=2tanθ,dx=2sec2θdθ,x2+4=4sec2θx = 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=2sec2θdθ4sec2θ\int \frac{dx}{x^2 + 4} = \int \frac{2\sec^2 \theta \, d\theta}{4\sec^2 \theta}

Step 3: Simplify:

dxx2+4=12dθ\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=2tanθx = 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

infoNote

Common Mistakes:

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

  2. Incorrect trigonometric identities: Verify sin2θ+cos2θ=1,sec2θ1=tan2θ\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.

infoNote

Key Formulas:

  1. Trigonometric Substitutions:
  • a2x2:x=asinθ,dx=acosθdθ\sqrt{a^2 - x^2}: x = a\sin \theta, dx = a\cos \theta \, d\theta
  • x2a2:x=asecθ,dx=asecθtanθdθ\sqrt{x^2 - a^2}: x = a\sec \theta, dx = a\sec \theta \tan \theta \, d\theta
  • x2+a2:x=atanθ,dx=asec2θdθ\sqrt{x^2 + a^2}: x = a\tan \theta, dx = a\sec^2 \theta \, d\theta
  1. Standard Integrals:
  • dxa2x2=arcsin(xa)+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C
  • dxx2a2=lnx+x2a2+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
Books

Only available for registered users.

Sign up now to view the full note, or log in if you already have an account!

500K+ Students Use These Powerful Tools to Master Integration by Substitution

Enhance your understanding with flashcards, quizzes, and exams—designed to help you grasp key concepts, reinforce learning, and master any topic with confidence!

50 flashcards

Flashcards on Integration by Substitution

Revise key concepts with interactive flashcards.

Try Further Maths Core Pure Flashcards

5 quizzes

Quizzes on Integration by Substitution

Test your knowledge with fun and engaging quizzes.

Try Further Maths Core Pure Quizzes

29 questions

Exam questions on Integration by Substitution

Boost your confidence with real exam questions.

Try Further Maths Core Pure Questions

27 exams created

Exam Builder on Integration by Substitution

Create custom exams across topics for better practice!

Try Further Maths Core Pure exam builder

50 papers

Past Papers on Integration by Substitution

Practice past papers to reinforce exam experience.

Try Further Maths Core Pure Past Papers

Other Revision Notes related to Integration by Substitution you should explore

Discover More Revision Notes Related to Integration by Substitution to Deepen Your Understanding and Improve Your Mastery

96%

114 rated

Methods in Calculus

Improper Integrals

user avatar
user avatar
user avatar
user avatar
user avatar

213+ studying

181KViews

96%

114 rated

Methods in Calculus

Mean Value of a Function

user avatar
user avatar
user avatar
user avatar
user avatar

481+ studying

189KViews

96%

114 rated

Methods in Calculus

Integrating with Partial Fractions

user avatar
user avatar
user avatar
user avatar
user avatar

273+ studying

192KViews

96%

114 rated

Methods in Calculus

Calculus Involving Inverse Trig

user avatar
user avatar
user avatar
user avatar
user avatar

243+ studying

188KViews
Load more notes

Join 500,000+ A-Level students using SimpleStudy...

Join Thousands of A-Level Students Using SimpleStudy to Learn Smarter, Stay Organized, and Boost Their Grades with Confidence!

97% of Students

Report Improved Results

98% of Students

Recommend to friends

500,000+

Students Supported

50 Million+

Questions answered