Interesting way to evaluate $ \int \cos^3 x\ dx$

Solution 1:

Maybe you'll like this method: \begin{eqnarray} \int\cos^3xdx&=&\int\cos^2xd\sin x\\ &=&\int(1-\sin^2x)d\sin x\\ &=&\sin x-\frac13\sin^3x+C. \end{eqnarray}

Solution 2:

Euler's approach:

\begin{align}\int\cos^3(x)~dx&=\int\left[\frac{e^{ix}+e^{-ix}}2\right]^3~dx\\&=\int\frac{e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}}8~dx\end{align}

At this point, either integrate directly, and then notice the resulting sine functions, or notice that

$$\frac{e^{3ix}+e^{-3ix}}8=\frac14\cos(3x)\\\frac{3e^{ix}+3e^{-ix}}8=\frac34\cos(x)$$

And thus the rest is simple.

Solution 3:

Use that $\cos 3x = 4\cos^3 x - 3\cos x$ to get:

$$\int \cos^3 x \,dx = \frac{1}{4}\int\left(\cos 3x +3\cos x\right)\,dx$$

and the right side is easy to compute as $\frac{1}{12}\sin 3x +\frac{3}{4}\sin x$.

My original answer included a second approach, but I had the wrong formula for $dx$ in the substitution.

Use the tangent-half-angle substitution, $t=\tan(x/2)$ so $dx = \frac{2}{1+t^2}\,dt$ (corrected) and $\cos(x)=\frac{1-t^2}{1+t^2}$.

This reduces to:

$$\int \frac{2(1-t^2)^3}{(1+t^2)^4}\,dt$$

But this is a bit harder to integrate than the nice formula I had before. You might still solve this if you rewrite $1-t^2=2-(1+t^2)$ and expand, and then compute:

$$\int \frac{dt}{\left(1+t^2\right)^k}$$ for $k=1,2,3,4.$