Odd power, peel and substitute: \displaystyle\int \sin^3 x\, dx
Step 1 — peel one \sin x off the odd power. Keep a
single \sin x aside (it will become du),
leaving an even power:
\sin^3 x = \sin^2 x \cdot \sin x.
Step 2 — convert the even part to cosines. Use
\sin^2 x = 1 - \cos^2 x, so everything but the lone
\sin x is in \cos x:
\int \sin^3 x\, dx = \int \big(1 - \cos^2 x\big)\sin x\, dx.
Step 3 — substitute u = \cos x. Then
du = -\sin x\, dx, so \sin x\, dx = -du:
\int \big(1 - \cos^2 x\big)\sin x\, dx = \int (1 - u^2)(-du) = -\int (1 - u^2)\, du.
Step 4 — integrate the polynomial in u.
-\int (1 - u^2)\, du = -\left(u - \frac{u^3}{3}\right) + C = -u + \frac{u^3}{3} + C.
Step 5 — back-substitute u = \cos x.
\int \sin^3 x\, dx = -\cos x + \frac{\cos^3 x}{3} + C.
Even power, reduce first: \displaystyle\int \cos^2 x\, dx
Here there is nothing to peel — \cos^2 x is even, and pulling out a
\cos x would leave an awkward odd power. Instead, halve the power.
Step 1 — apply the power-reduction identity.
\cos^2 x = \frac{1 + \cos 2x}{2}.
Step 2 — substitute it into the integral. The integrand is now a sum of
things we can integrate directly:
\int \cos^2 x\, dx = \int \frac{1 + \cos 2x}{2}\, dx = \frac{1}{2}\int 1\, dx + \frac{1}{2}\int \cos 2x\, dx.
Step 3 — integrate each piece. The first is x; for
the second, \int \cos 2x\, dx = \tfrac{1}{2}\sin 2x (a tiny
substitution u = 2x):
\frac{1}{2}\int 1\, dx + \frac{1}{2}\int \cos 2x\, dx = \frac{x}{2} + \frac{1}{2}\cdot\frac{\sin 2x}{2}.
Step 4 — collect.
\int \cos^2 x\, dx = \frac{x}{2} + \frac{\sin 2x}{4} + C.
To evaluate \displaystyle\int \sin^m x\, \cos^n x\, dx:
-
If m (the power of sine) is odd: split off one
\sin x, rewrite the rest with
\sin^2 x = 1 - \cos^2 x, and substitute
u = \cos x.
-
If n (the power of cosine) is odd: split off one
\cos x, rewrite with
\cos^2 x = 1 - \sin^2 x, and substitute
u = \sin x.
-
If both are even: lower the powers with the power-reduction identities
\sin^2 x = \tfrac{1 - \cos 2x}{2} and
\cos^2 x = \tfrac{1 + \cos 2x}{2}, then integrate term by term.
The identities that do all the work:
\sin^2 x + \cos^2 x = 1, \qquad \sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2}.
For the secant/tangent family, the Pythagorean identity in its other guise is the key:
\sec^2 x = 1 + \tan^2 x, \qquad \frac{d}{dx}\tan x = \sec^2 x, \qquad \frac{d}{dx}\sec x = \sec x \tan x.
The famous \int \sec x\, dx trick. Multiply and
divide by \sec x + \tan x:
\int \sec x\, dx = \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x}\, dx = \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x}\, dx.
The numerator is exactly the derivative of the denominator, so the substitution
u = \sec x + \tan x turns it into
\int \tfrac{du}{u} = \ln|u|:
\int \sec x\, dx = \ln\big|\sec x + \tan x\big| + C.