Evaluating $\int P(\sin x, \cos x) \text{d}x$

There are several general approaches to integrals that involve expressions with sines and cosines, polynomial expressions being the simplest ones.

Weierstrass Substitution

A method that always works is Weierstrass substitution, which will turn such an integral into an integral of rational functions, which in turn can always be solved, at least in principle, by the method of partial fractions. This works even for rational functions of sine and cosine, as well as functions that involve the other trigonometric functions.

Weierstrass substitution replaces sines and cosines (and by extension, tangents, cotangents, secants, and cosecants) by rational functions of a new variable. The identities begin by the trigonometric substitution $t = \tan\frac{x}{2}$, with $-\pi\lt x\lt \pi$, which yields $$\begin{align*} \sin x &= \frac{2t}{1+t^2}\\ \cos x &= \frac{1-t^2}{1+t^2}\\ dx &= \frac{2\,dt}{1+t^2}. \end{align*}$$ For example, if we have $$\int \frac{\sin x-\cos x}{\sin x+\cos x}\,dx$$ using the substitution above we obtain: $$\begin{align*} \int\frac{\sin x-\cos x}{\sin x+\cos x}\,dx &= \int\left(\frac{\quad\frac{2t}{1+t^2} - \frac{1-t^2}{1+t^2}\quad}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}}\right)\left(\frac{2}{1+t^2}\right)\,dt\\ &= \int\left(\frac{\quad\frac{2t-1+t^2}{1+t^2}\quad}{\frac{1+2t-t^2}{1+t^2}}\right) \left(\frac{2}{1+t^2}\right)\,dt\\ &= \int\left(\frac{2t-1+t^2}{2t+1-t^2}\right)\left(\frac{2}{1+t^2}\right)\,dt\\ &= 2\int\frac{2t-1+t^2}{(1+t^2)(2t+1-t^2)}\,dt \end{align*}$$ which can then be integrated by the method of partial fractions.

Substitutions and Reduction formulas

However, there are usually faster methods, particularly for polynomial expressions. By breaking up the integral into a sum of integrals corresponding to the monomials, the problem reduces to solving integrals of the form $$\int \left(\sin x\right)^n \left(\cos x\right)^m\,dx$$ with $n$ and $m$ nonnegative integers. The standard methods then are:

1. If $n$ is odd, then "reserve" one sine, and transform the others into cosines by using the identity $\sin^2 x = 1-\cos^2x$. Then do the change of variable $u=\cos x$ to transform the integral into the integral of a polynomial. For example, $$\int \left(\sin x\right)^5\left(\cos x\right)^2\,dx,$$ then take $(\sin x)^5$, and write it as $$\sin x(\sin x)^4 = \sin x(\sin^2x)^2 = \sin x(1-\cos^2 x)^2.$$ Then setting $u=\cos x$ and $du = -\sin x\,dx$, we get $$\int\left(\sin x\right)^4\left(\cos x\right)^2\,dx = \int \sin x\left(1-\cos^2x\right)^2\left(\cos x\right)^2\,dx = -\int (1-u^2)^2u^2\,du,$$ which can be solved easily.

2. If $m$ is odd, then do the same trick by by "reserving" one cosine and using the substitution $u=\sin x$. For example, $$\int \sin^2x\cos^3x\,dx = \int \sin^2x(\cos^2x)\cos x\,dx = \int(\sin^2x)(1-\sin^2x)\cos x\,dx$$ and then setting $u=\sin x$, $du = \cos x\,dx$, we get $$\int \sin^2x\cos^3x\,dx = \int u^2(1-u^2)\,du,$$ which can be solved easily again.

3. If $n$ and $m$ are both even, then either replace all the sines with cosines or vice versa, using $\sin^2 x = 1 - \cos^2x$ or $\cos^2x = 1-\sin^2 x$, and expand. This will leave integrals of the form $$\int \sin^n x\,dx\qquad\text{or}\quad \int \cos^m x\,dx$$ with $n$ and $m$ even positive and even. In that situation, one can use the reduction formulas, which can be obtained by using integration by parts: $$\begin{align*} \int \sin^n x\,dx &= - \frac{1}{n}\sin^{n-1} x\cos x + \frac{n-1}{n}\int \sin^{n-2}x\,dx,\\ \int \cos^m x\,dx &= \frac{1}{m}\cos^{m-1} x\sin x + \frac{n-1}{n}\int \cos^{n-2}x\,dx. \end{align*}$$ By repeated application of these formulas, one eventually ends up with an integral of the form $\int \,dx$ which can be solved directly.

The process can be shortened if you happen to spot or know some trigonometric identities; for example, the power reduction formulas allow you to replace powers of sines or cosines by expressions of multiple angles, e.g., $$\sin^4\theta = \frac{3-4\cos(2\theta)+\cos(4\theta)}{8}$$ could replace a single integral with three integrals that can be done fairly easily via substitution.

Other methods

If you are comfortable enough integrating functions with a complex variable in it, then the method described by Qiaochu will transform any integral that involves sines and cosines in any way into an integral that involves exponential functions instead.

Calculus books teach an annoying method based around using trig identities to reduce the integral to one where trig substitution can be applied. This method requires a little bit of guess-work to determine which identity should be applied, and my recollection is that it does not always work.

Here is a completely mechanical method which always works, although for simple $P$ it may require more calculation than a smarter method. Instead of using $\sin \theta, \cos \theta$, use the complex exponential $e^{i \theta}$; then Euler's formula $e^{i \theta} = \cos \theta + i \sin \theta$ tells you that $$\cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}, \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}$$

and now the problem is reduced to integrating a sum of exponentials.

Example. The integral $\int_{0}^{2\pi} \cos^{2n} \theta \, d \theta$ has come up several times on math.SE in one form or another. It is readily solved using this method: write it as

$$\frac{1}{4^n} \int_0^{2\pi} (e^{i \theta} + e^{-i \theta})^{2n} \, d \theta$$

and note that when integrating from $0$ to $2\pi$ all of the terms vanish except the constant term, so the final answer is

$$\frac{2\pi}{4^n} {2n \choose n}.$$

This method generalizes to the case when $P$ is replaced by a rational function; in that case the integrand becomes a rational function of $e^{i \theta}$ (rather than a Laurent polynomial) and using $u$-substitution we can reduce the problem to integrating a rational function, which can be done in a number of ways (partial fractions, residues...).

Here are some other substitutions that you can try on a rational function of trigonometric functions. We name them Bioche substitution in France.

Let $P(\sin t,\cos t)=f(t)$ where $P(x,y)$ is rational function. Let $\omega(t)=f(t)dt$.

• If $\omega(-t)=\omega(t)$, then $u(t)=\cos t$ might be a good substitution.
  For example : $$\int \frac{\sin^3t}{2+\cos t}dt=-\int \frac{(1-\cos^2t)(-\sin t)}{2+\cos t}dt=-\int\frac{1-u^2}{2+u}=\int\frac{u^2-1}{2+u}=\int u-2+\frac{3}{u-2}du=\int u du-2\int du+3\int \frac{1}{u-2}=\frac{u^2}{2}-2u+3\log(u-2)$$
• If $\omega(\pi-t)=\omega(t)$, then $u(t)=\sin t$ might be a good substitution.
  For example : $$\int \frac{1}{\cos t}dt=\int \frac{\cos t}{\cos^2 t}dt=\int \frac{\cos t}{1-\sin^2 t}dt=\int \frac{1}{1-u^2}du=\int \frac{1}{2} \bigg(\frac{1}{1+u}+\frac{1}{1-u}\bigg)du=\frac{1}{2}(\log(u+1)-\log(1-u))$$
• If $\omega(\pi+t)=\omega(t)$, then $u(t)=\tan t$ might be a good substitution.
  For example : $$\int\frac{1}{1+\cos^2 t}dt=\int \frac{1}{1+\frac{1}{\cos^2 t}}\frac{dt}{\cos^2 t}=\int \frac{1}{1+\frac{\cos^2t+\sin^2t}{\cos^2 t}}\frac{dt}{\cos^2 t}=\int \frac{1}{2+\tan^2t}\frac{dt}{\cos^2 t}=\int\frac{1}{2+u^2}du=\frac{1}{\sqrt2}\arctan\frac{u}{\sqrt2}$$
• If two of the previous relations are verified (in this case the three relations are verified), then $u=\cos(2t)$ might be a good substitution.

If none of these work, you can use the Weierstrass substitution presented in a previous answer.