Trigonometric Integrals

Friday: Quiz 2
Next: Trig subst.

$$\cos^2(x) =\frac{1+\cos(2x)}{2} \qquad \sin^2(x) = \frac{1-\sin(2x)}{2}.$$ (5.2)

Example 5.2.1   Compute $\int \sin^3(x)dx$.
We use trig. identities and compute the integral directly as follows:

$$\int \sin^3(x)dx = \int \sin^2(x) \sin(x) dx$$

$$= \int [1-\cos^2(x)]\sin(x) dx$$

$$= -\cos(x) + \frac{1}{3}\cos^3(x) + c u=\cos(x)$$

This always works for odd powers of $\sin(x)$.

Example 5.2.2   What about even powers?! Compute $\int \sin^4(x)dx$. We have

$$\sin^4(x) = [\sin^2(x)]^2$$

$$= \left[\frac{1-\cos(2x)}{2}\right]^2$$

$$= \frac{1}{4}\cdot \left[ 1 - 2\cos(2x) + \cos^2(2x)\right]$$

$$= \frac{1}{4}\left[ 1 - 2\cos(2x) + \frac{1}{2} + \frac{1}{2} \cos(4x) \right]$$

Thus

$$\int \sin^4(x) dx = \int \left[\frac{3}{8} - \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right] dx$$

$$= \frac{3}{8} x - \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x) + c.$$

Key Trick: Realize that we should write $\sin^4(x)$ as $(\sin^2(x))^2$. The rest is straightforward.

Example 5.2.3   This example illustrates a method for computing integrals of trig functions that doesn't require knowing any trig identities at all or any tricks. It is very tedious though. We compute $\int \sin^3(x)dx$ using complex exponentials. We have

$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2} \qquad \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}.$$

hence

$$\int \sin^3(x)dx = \int \left(\frac{e^{ix} - e^{-ix}}{2i}\right)^3 dx$$

$$= -\frac{1}{8i} \int (e^{ix} - e^{-ix})^3 dx$$

$$= -\frac{1}{8i} \int (e^{ix} - e^{-ix})(e^{ix} - e^{-ix})(e^{ix} - e^{-ix})dx$$

$$= -\frac{1}{8i} \int (e^{2ix} -2 + e^{-2ix})(e^{ix} - e^{-ix})dx$$

$$= -\frac{1}{8i} \int e^{3ix} - e^{ix} - 2e^{ix} + 2e^{-ix} + e^{-ix} - e^{-3ix} dx$$

$$= -\frac{1}{8i} \int e^{3ix}- e^{-3ix} + 3e^{-ix} - 3e^{ix} dx$$

$$= -\frac{1}{8i} \left(\frac{e^{3ix}}{3i} - \frac{e^{-3ix}}{-3i} + \frac{3e^{-ix}}{-i} - \frac{3e^{ix}}{i}\right) + c$$

$$= \frac{1}{4} \left( \frac{1}{3} \cos(3x) - 3\cos(x)\right) + c$$

$$= \frac{1}{12} \cos(3x) - \frac{3}{4}\cos(x) + c$$

The answer looks totally different, but is in fact the same function.

Here are some more identities that we'll use in illustrating some tricks below.

$$\frac{d}{dx} \tan(x) = \sec^2(x)$$

and

$$\frac{d}{dx} \sec(x) = \sec(x)\tan(x).$$

Also,

$$1 + \tan^2(x) = \sec^2(x).$$

Example 5.2.4   Compute $\int \tan^3(x) dx$. We have

$$\int \tan^3(x) dx = \int\tan(x) \tan^2(x) dx$$

$$= \int \tan(x)\left[\sec^2(x) - 1\right]dx$$

$$= \int \tan(x)\sec^2(x)dx - \int\tan(x) dx$$

$$= \frac{1}{2} \tan^2(x) - \ln\vert\sec(x)\vert + c$$

Here we used the substitution $u=\tan(x)$, so $du = \sec^2(x) dx$, so

$$\int \tan(x)\sec^2(x)dx = \int u du = \frac{1}{2}u^2 + c = \frac{1}{2}\tan^2(x) + c.$$

Also, with the substitution $u=\cos(x)$ and $du=-\sin(x)dx$ we get

$$\int \tan(x)dx = \int \frac{\sin(x)}{\cos(x)} dx = -\int \frac{1}{u} du = -\ln\vert u\vert + c = -\ln\vert\sec(x)\vert + c.$$

Key trick: Write $\tan^3(x)$ as $\tan(x)\tan^2(x)$.

Example 5.2.5   Here's one that combines trig identities with the funnest variant of integration by parts. Compute $\int \sec^3(x) dx$.
We have

$$\int \sec^3(x)dx = \int \sec(x) \sec^2(x)dx.$$

Let's use integration by parts.

$$u = \sec(x) v = \tan(x)$$

$$du = \sec(x)\tan(x) dx dv = \sec^2(x)dx$$

The above integral becomes

$$\int \sec(x) \sec^2(x)dx = \sec(x) \tan(x) - \int \sec(x) \tan^2(x) dx$$

$$=\sec(x) \tan(x) - \int \sec(x)[\sec^2(x) - 1] dx$$

$$= \sec(x) \tan(x) - \int \sec^3(x) + \int \sec(x) dx$$

$$= \sec(x) \tan(x) - \int \sec^3(x) + \ln\vert\sec(x) + \tan(x)\vert$$

This is familiar. Solve for $\int \sec^3(x)$. We get

$$\int \sec^3(x) dx = \frac{1}{2}\Bigl[ \sec(x) \tan(x) + \ln\vert\sec(x) + \tan(x)\vert \Bigr] + c$$