Integration Using Trigonometric Identities

The idea

Functions like $\cos^2 x$ and $\sin^2 x$ have no direct antiderivative, but the double-angle identities rewrite them as something that does:

\cos^2 x = \dfrac{1+\cos 2x}{2}, \qquad \sin^2 x = \dfrac{1-\cos 2x}{2}

Power-reduction and product-to-sum identities, and what each one integrates

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

Solution: Substituting the identities above and integrating term by term:

\int \cos^2 x\,dx = \int \dfrac{1+\cos 2x}{2}\,dx = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C

\int \sin^2 x\,dx = \int \dfrac{1-\cos 2x}{2}\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C

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

Solution: Use $2\sin A\cos B = \sin(A + B) + \sin(A - B)$ with $A = 3x$ and $B = 2x$ :

\int \sin 3x\cos 2x\,dx = \dfrac{1}{2}\int (\sin 5x + \sin x)\,dx = -\dfrac{\cos 5x}{10} - \dfrac{\cos x}{2} + C

Find $\displaystyle\int \tan^2 x\,dx$ .

Solution: Replace $\tan^2 x$ with $\sec^2 x - 1$ , both of which have standard antiderivatives:

\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x + C

Find $\displaystyle\int \dfrac{1-\sin x}{\cos^2 x}\,dx$ .

Solution: Split the fraction into two terms:

\int \dfrac{1}{\cos^2 x}\,dx - \int \dfrac{\sin x}{\cos^2 x}\,dx = \int \sec^2 x\,dx - \int \sec x\tan x\,dx

Both are standard forms:

= \tan x - \sec x + C

Dropping the factor from $\cos 2x$ . $\int \cos 2x\,dx = \dfrac{\sin 2x}{2}$ , not $\sin 2x$ . The same goes for $\sin 5x$ , $\cos 3x$ and so on.
Getting the order wrong in $2\sin A\sin B$ . It is $\cos(A - B) - \cos(A + B)$ . Reversing it flips the sign.
Integrating $\sin^2 x$ as $\dfrac{\sin^3 x}{3}$ . The power rule only works for a power of $x$ itself. Use the power-reduction identity.