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}

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

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