Integrating sin² x and cos² x

Using the double angle formulas to integrate sin²x and cos²x, integrals such as sin²(ax), the area of a circle by the substitution x = sin θ, and a summary of the standard integrals.

Squares of sine and cosine

From the double angle formula

$$\sin^2 x = \tfrac12(1 - \cos 2x) \qquad \cos^2 x = \tfrac12(1 + \cos 2x)$$

$$\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C \qquad \int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$$

Example 1●●

Evaluate $\displaystyle\int_0^{\frac{\pi}{2}} \sin^2 x\,dx$ and $\displaystyle\int_0^{\frac{\pi}{4}} \cos^2 x\,dx$.

$\Big[\tfrac{x}{2} - \tfrac{\sin 2x}{4}\Big]_0^{\frac{\pi}{2}} = \tfrac{\pi}{4} - 0 = \tfrac{\pi}{4}$.

$\Big[\tfrac{x}{2} + \tfrac{\sin 2x}{4}\Big]_0^{\frac{\pi}{4}} = \tfrac{\pi}{8} + \tfrac14$.

Example 2●●

Find $\displaystyle\int \sin^2 3x\,dx$.

$\sin^2 3x = \tfrac12(1 - \cos 6x)$, so the integral is $\dfrac{x}{2} - \dfrac{\sin 6x}{12} + C$.

A trigonometric substitution

Example 3: a quarter circle●●●

Use $x = \sin\theta$ to evaluate $\displaystyle\int_0^1 \sqrt{1 - x^2}\,dx$.

$dx = \cos\theta\,d\theta$. The limits become $\theta = 0$ and $\theta = \frac{\pi}{2}$, and $\sqrt{1 - \sin^2\theta} = \cos\theta$ there.

$\displaystyle\int_0^{\frac{\pi}{2}} \cos^2\theta\,d\theta = \Big[\tfrac{\theta}{2} + \tfrac{\sin 2\theta}{4}\Big]_0^{\frac{\pi}{2}} = \tfrac{\pi}{4}$.

This is the area of a quarter of the unit circle, as it should be.

Summary of the standard integrals

$\displaystyle\int \sin ax\,dx = -\tfrac1a\cos ax + C$	$\displaystyle\int \cos ax\,dx = \tfrac1a\sin ax + C$
$\displaystyle\int \sec^2 ax\,dx = \tfrac1a\tan ax + C$	$\displaystyle\int \frac{dx}{a^2 + x^2} = \tfrac1a\tan^{-1}\tfrac{x}{a} + C$
$\displaystyle\int \sin^2 x\,dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C$	$\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\tfrac{x}{a} + C$

Flashcards

$sin^2 x$ in terms of $cos 2x$

$\tfrac12(1 - \cos 2x)$

$cos^2 x$ in terms of $cos 2x$

$\tfrac12(1 + \cos 2x)$

$displaystyleint cos^2 x,dx$

$\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$

Practice

Evaluate $\displaystyle\int_0^{\frac{\pi}{2}} \sin^2 x\,dx$. Type pi for $\pi$.
Evaluate $\displaystyle\int_0^{\frac{\pi}{4}} \cos^2 x\,dx$.
Evaluate $\displaystyle\int_0^{\pi} \sin^2 x\,dx$.
Evaluate $\displaystyle\int_0^1 \sqrt{1 - x^2}\,dx$.
Evaluate $\displaystyle\int_0^{\frac{\pi}{2}} \cos^2 2x\,dx$.
Evaluate $\displaystyle\int_0^{2\pi} \cos^2 x\,dx$.
$\cos^2 x = $
$\displaystyle\int \sin^2 3x\,dx = $
The substitution $x = \sin\theta$ turns $\sqrt{1 - x^2}$ into: