Definite Integrals of Trigonometry by Substitution (Exercise 7.9)

This lesson evaluates a set of definite integrals of trigonometric functions from Exercise 7.9 using the substitution method. The key idea throughout is that when you replace $x$ with a new variable, you also convert the limits of integration to that variable, so you never have to substitute back.

A simple power of sine

Evaluate

$I = \int_0^{\pi/4} \sin^2 2x \, \cos 2x \, dx.$

Put $t = \sin 2x$. Then $dt = 2\cos 2x \, dx$, so $\cos 2x \, dx = \tfrac{1}{2}\,dt$.

Change the limits. When $x = 0$, $t = \sin 0 = 0$. When $x = \tfrac{\pi}{4}$, $t = \sin\tfrac{\pi}{2} = 1$.

$I = \frac{1}{2}\int_0^1 t^2 \, dt = \frac{1}{2}\left[\frac{t^3}{3}\right]_0^1 = \frac{1}{6}\big(1^3 - 0^3\big) = \frac{1}{6}.$

Inverse tangent over $1 + x^2$

Evaluate

$I = \int_0^1 \frac{\tan^{-1} x}{1 + x^2}\, dx.$

Put $t = \tan^{-1} x$, so $dt = \dfrac{1}{1 + x^2}\,dx$.

Change the limits. When $x = 0$, $t = \tan^{-1} 0 = 0$. When $x = 1$, $t = \tan^{-1} 1 = \tfrac{\pi}{4}$.

$I = \int_0^{\pi/4} t \, dt = \left[\frac{t^2}{2}\right]_0^{\pi/4} = \frac{1}{2}\left(\frac{\pi}{4}\right)^2 = \frac{1}{2}\cdot\frac{\pi^2}{16} = \frac{\pi^2}{32}.$

Root sine times cosine to the fifth power

Evaluate

$I = \int_0^{\pi/2} \sqrt{\sin\phi}\,\cos^5\phi \, d\phi.$

Split off one cosine and rewrite the rest with $\cos^4\phi = (1 - \sin^2\phi)^2$:

$I = \int_0^{\pi/2} \sqrt{\sin\phi}\,\big(1 - \sin^2\phi\big)^2 \cos\phi \, d\phi.$

Put $t = \sin\phi$, so $dt = \cos\phi \, d\phi$. The limits become $0$ and $1$.

$I = \int_0^1 \sqrt{t}\,(1 - t^2)^2 \, dt = \int_0^1 t^{1/2}\big(1 - 2t^2 + t^4\big)\, dt = \int_0^1 \Big(t^{1/2} - 2t^{5/2} + t^{9/2}\Big)\, dt.$

Integrating term by term:

$I = \left[\frac{t^{3/2}}{3/2} - 2\cdot\frac{t^{7/2}}{7/2} + \frac{t^{11/2}}{11/2}\right]_0^1 = \frac{2}{3} - \frac{4}{7} + \frac{2}{11}.$

With a common denominator of $231$:

$I = \frac{154 - 132 + 42}{231} = \frac{64}{231}.$

Inverse sine substitution with integration by parts

Evaluate

$I = \int_0^1 \sin^{-1}\!\left(\frac{2x}{1 + x^2}\right) dx.$

Put $x = \tan\theta$, so $dx = \sec^2\theta \, d\theta$. Then

$\frac{2x}{1 + x^2} = \frac{2\tan\theta}{1 + \tan^2\theta} = \sin 2\theta,$

so $\sin^{-1}(\sin 2\theta) = 2\theta$. The limits $x = 0$ and $x = 1$ become $\theta = 0$ and $\theta = \tfrac{\pi}{4}$.

$I = \int_0^{\pi/4} 2\theta \, \sec^2\theta \, d\theta = 2\int_0^{\pi/4} \theta \, \sec^2\theta \, d\theta.$

Integration by parts with first function $\theta$ and second function $\sec^2\theta$ (whose integral is $\tan\theta$):

$I = 2\left[\theta \tan\theta\right]_0^{\pi/4} - 2\int_0^{\pi/4} \tan\theta \, d\theta = 2\left[\theta \tan\theta\right]_0^{\pi/4} - 2\Big[\log|\sec\theta|\Big]_0^{\pi/4}.$

Evaluating, $\tan\tfrac{\pi}{4} = 1$ and $\sec\tfrac{\pi}{4} = \sqrt{2}$:

$I = 2\cdot\frac{\pi}{4}\cdot 1 - 2\log\sqrt{2} = \frac{\pi}{2} - 2\cdot\tfrac{1}{2}\log 2 = \frac{\pi}{2} - \log 2.$

Sine over $1 + \cos^2 x$

Evaluate

$I = \int_0^{\pi/2} \frac{\sin x}{1 + \cos^2 x}\, dx.$

Put $t = \cos x$, so $-\sin x \, dx = dt$, that is $\sin x \, dx = -dt$.

Change the limits. When $x = 0$, $t = \cos 0 = 1$. When $x = \tfrac{\pi}{2}$, $t = \cos\tfrac{\pi}{2} = 0$.

$I = \int_1^0 \frac{-dt}{1 + t^2} = \int_0^1 \frac{dt}{1 + t^2} = \Big[\tan^{-1} t\Big]_0^1 = \tan^{-1} 1 - \tan^{-1} 0 = \frac{\pi}{4}.$

Reading a derivative straight from an integral

If

$f(x) = \int_0^x t \sin t \, dt,$

then by the fundamental theorem of calculus, differentiating an integral with respect to its upper limit gives back the integrand:

$f'(x) = x \sin x.$

Key takeaways

• When you substitute a new variable, convert the limits to that variable so you can evaluate without changing back.
• For powers of sine and cosine, peel off one factor to match the differential and rewrite the rest with $\sin^2\phi + \cos^2\phi = 1$.
• An inverse-trig substitution like $x = \tan\theta$ can turn an awkward inverse-sine integral into a simple integration by parts.
• The derivative of $\int_0^x g(t)\,dt$ with respect to $x$ is just $g(x)$.