Trigonometric Substitution for Beginners: A Step-by-Step Guide

Step 4: Solving the Integral

Integrate the resulting trigonometric integral.

Step 5: Converting Back to x

Use a right triangle to express \( \theta \) in terms of x.

Worked Out Examples

For each case, we will work through an indefinite integral and a definite integral.

Case 1: \( \sqrt{b^2 – a^2x^2} \Rightarrow x = \frac{b}{a}\sin\theta \)

Example 1: Evaluate \( \int \frac{x^2}{\sqrt{1 – x^2}} dx \).

Solution: Let \( x = \sin{\theta} \), then \( dx = \cos{\theta} d\theta \). Substituting into the integral, we get

$$ \int \frac{\sin^2{\theta}}{\sqrt{1 – sin^2{\theta}}}\cos{\theta} d\theta = \int \frac{\sin^2{\theta}}{\sqrt{cos^2{\theta}}}\cos{\theta} d\theta = \int \frac{\sin^2{\theta}}{\cos{\theta}}\cos{\theta} d\theta = \int \sin^2{\theta} d\theta. $$

Applying the power reducing identity gives

$$ \int \frac{1}{2} – \frac{1}{2}\cos{2\theta} d\theta = \frac{1}{2}\theta – \frac{1}{4}\sin{2\theta} + C = \frac{1}{2}\theta – \frac{1}{2}\sin{\theta}\cos{\theta} + C. $$

Using the triangle to convert everything back into terms of x, we arrive at

$$ \frac{1}{2}\arcsin{\theta} – \frac{1}{2}x\sqrt{1 – x^2} + C. $$

Example 2: Evaluate \( \int_0^1 \frac{x^2}{\sqrt{1 – x^2}} dx \).

Solution: Let \( x = \sin{\theta} \), then \( dx = \cos{\theta} ol\theta \). By substituting the limits into the substitution and solving the resulting trigonometric equations, the new limits are \( 0 \Rightarrow 0 \) and \( 1 \Rightarrow \frac{\pi}{2} \). Substituting into the integral, we get

$$ \int_0^{\frac{\pi}{2}} \frac{\sin^2{\theta}}{\sqrt{1 – sin^2{\theta}}}\cos{\theta} d\theta = \int_0^{\frac{\pi}{2}} \frac{\sin^2{\theta}}{\sqrt{cos^2{\theta}}}\cos{\theta} d\theta = \int_0^{\frac{\pi}{2}} \frac{\sin^2{\theta}}{|\cos{\theta}|}\cos{\theta} d\theta = \int_0^{\frac{\pi}{2}} \frac{\sin^2{\theta}}{\cos{\theta}}\cos{\theta} d\theta = \int_0^{\frac{\pi}{2}} \sin^2{\theta} d\theta. $$

Applying the power reducing identity gives

$$ \int_0^{\frac{\pi}{2}} \frac{1}{2} – \frac{1}{2}\cos{2\theta} d\theta = (\frac{1}{2}\theta – \frac{1}{4}\sin{2\theta})|_0^{\frac{\pi}{2}} = \frac{\pi}{4}. $$

Case 2: \( \sqrt{a^2x^2 + b^2} \Rightarrow x = \frac{b}{a}\tan\theta \)

Example 3: Evaluate \( \int \frac{1}{\sqrt{x^2 + 1}} dx \).

Solution: Let \( x = \tan{\theta} \), then \( dx = \sec^2{\theta} d\theta \). Substituting into the integral, we get

$$ \int \frac{1}{\sqrt{tan^2{\theta} + 1}}\sec^2{\theta} d\theta = \int \frac{1}{\sqrt{sec^2{\theta}}}\sec^2{\theta} d\theta = \int \frac{1}{\sec{\theta}}\sec^2{\theta} d\theta = \int \sec{\theta} d\theta = \int \frac{\sec{\theta}(\sec{\theta} + \tan{\theta})}{\sec{\theta} + \tan{\theta}} d\theta = \int \frac{\sec^2{\theta} + \sec{\theta}\tan{\theta}}{\sec{\theta} + \tan{\theta}} d\theta. $$

Let \( u = \sec{\theta} + \tan{\theta} \), then \( du = \sec^2{\theta} + \sec{\theta}\tan{\theta} \). Substituting into the integral, we get

$$ \int \frac{1}{u} du = \ln{|u|} + C = \ln{|\sec{\theta} + \tan{\theta}|} + C. $$

Using the triangle to convert everything back into terms of x, we arrive at

$$ \ln{|x + \sqrt{x^2 + 1}|} + C. $$

Example 4: Evaluate \( \int_0^1 \frac{1}{\sqrt{x^2 + 1}} dx \).

Solution: Let \( x = \tan{\theta} \), then \( dx = \sec^2{\theta} d\theta \). By substituting the limits into the substitution and solving the resulting trigonometric equations, the new limits are \( 0 \Rightarrow 0 \) and \( 1 \Rightarrow \frac{\pi}{4} \). Substituting into the integral, we get

$$ \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{tan^2{\theta} + 1}}\sec^2{\theta} d\theta = \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{sec^2{\theta}}}\sec^2{\theta} d\theta = \int_0^{\frac{\pi}{4}} \frac{1}{|\sec{\theta}|}\sec^2{\theta} d\theta = \int_0^{\frac{\pi}{4}} \frac{1}{\sec{\theta}}\sec^2{\theta} d\theta = \int_0^{\frac{\pi}{4}} \sec{\theta} d\theta = \int_0^{\frac{\pi}{4}} \frac{\sec{\theta}(\sec{\theta} + \tan{\theta})}{\sec{\theta} + \tan{\theta}} d\theta = \int_0^{\frac{\pi}{4}} \frac{\sec^2{\theta} + \sec{\theta}\tan{\theta}}{\sec{\theta} + \tan{\theta}} d\theta. $$

Let \( u = \sec{\theta} + \tan{\theta} \), then \( u = \sec^2{\theta} + \sec{\theta}\tan{\theta} \). The new limits are \( 0 \Rightarrow 1 \) and \( \frac{\pi}{4} \Rightarrow 1 + \sqrt{2} \). Substituting into the integral we get

$$ \int_1^{1+ \sqrt{2}} \frac{1}{u} du = \ln{|u|}|_1^{1+ \sqrt{2}} = \ln{(1 + \sqrt{2})}. $$

Case 3: \( \sqrt{a^2x^2 – b^2} \Rightarrow x = \frac{b}{a}\sec\theta \)

Example 5: Evaluate \( \int \frac{1}{\sqrt{x^2 – 1}} dx \).

Solution: Let \( x = \sec{\theta} \), then \( dx = \sec{\theta}\tan{\theta} d\theta \). Substituting into the integral, we get

$$ \int \frac{1}{\sqrt{sec^2{\theta} – 1}}\sec{\theta}\tan{\theta} d\theta = \int \frac{1}{\sqrt{tan^2{\theta}}}\sec{\theta}\tan{\theta} d\theta = \int \frac{1}{\tan{\theta}}\sec{\theta}\tan{\theta} d\theta = \int \sec{\theta} d\theta = \int \frac{\sec{\theta}(\sec{\theta} + \tan{\theta})}{\sec{\theta} + \tan{\theta}} d\theta = \int \frac{\sec^2{\theta} + \sec{\theta}\tan{\theta}}{\sec{\theta} + \tan{\theta}} d\theta. $$

Let \( u = \sec{\theta} + \tan{\theta} \), then \( u = \sec^2{\theta} + \sec{\theta}\tan{\theta} \). Substituting into the integral, we get

$$ \int \frac{1}{u} du = \ln{|u|} + C = \ln{|\sec{\theta} + \tan{\theta}|} + C. $$

Using the triangle to convert everything back into terms of x, we arrive at

$$ \ln{|x + \sqrt{x^2 – 1}|} + C. $$

Example 6: Evaluate \( \int_1^2 \frac{1}{\sqrt{x^2 – 1}} dx \).

Solution: Let \( x = \sec{\theta} \), then \( dx = \sec{\theta}\tan{\theta} d\theta \). By substituting the limits into the substitution and solving the resulting trigonometric equations, the new limits are \( 1 \Rightarrow 0 \) and \( 2 \Rightarrow \frac{\pi}{3} \). Substituting into the integral, we get

$$ \int_0^{\frac{\pi}{3}} \frac{1}{\sqrt{sec^2{\theta} – 1}}\sec{\theta}\tan{\theta} d\theta = \int_0^{\frac{\pi}{3}} \frac{1}{\sqrt{tan^2{\theta}}}\sec{\theta}\tan{\theta} d\theta = \int_0^{\frac{\pi}{3}} \frac{1}{|\tan{\theta}|}\sec{\theta}\tan{\theta} d\theta = \int_0^{\frac{\pi}{3}} \frac{1}{\tan{\theta}}\sec{\theta}\tan{\theta} d\theta = \int_0^{\frac{\pi}{3}} \sec{\theta} d\theta = \int_0^{\frac{\pi}{3}} \frac{\sec{\theta}(\sec{\theta} + \tan{\theta})}{\sec{\theta} + \tan{\theta}} d\theta = \int_0^{\frac{\pi}{3}} \frac{\sec^2{\theta} + \sec{\theta}\tan{\theta}}{\sec{\theta} + \tan{\theta}} d\theta. $$

Let \( u = \sec{\theta} + \tan{\theta} \), then \( u = \sec^2{\theta} + \sec{\theta}\tan{\theta} \). The new limits are \( 0 \Rightarrow 1 \) and \( \frac{\pi}{3} \Rightarrow 2 + \sqrt{3} \). Substituting into the integral, we get

$$ \int_1^{2 + \sqrt{3}} \frac{1}{u} du = \ln{|u|}|_1^{2 + \sqrt{3}} = \ln{2 + \sqrt{3}}. $$

The last example shows that another method, like integration by parts, must sometimes be used before we can use a trigonometric substitution. For help with integration by parts, see my article Integration by Parts Explained with Examples: A Step-by-Step Guide.

Example 7: Using example 5 as a lemma, evaluate \( \int \text{arcsec}(x) dx \).

Solution: Choose \( u = \text{arcsec}(x) \) and \( dv = dx \), then \( du = \frac{1}{x\sqrt{x^2 – 1}}dx \) and \( v = x \). Applying the integration by parts formula then gives

$$ \int \text{arcsec}(x) \, dx = x \text{arcsec}(x) – \int \frac{1}{\sqrt{x^2 – 1}} dx = x \text{arcsec}(x) – \ln{|x + \sqrt{x^2 – 1}|} + C.$$

Conclusion

Trigonometric substitution is an important technique in integral calculus. By recognizing the appropriate quadratics and corresponding substitutions, these more specialized integrals can be evaluated. Practice this method to become more proficient in using trigonometric substitution.

Frequently Asked Questions