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

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}

\begin{eqnarray*} \int \cos^3(x) dx &=& \int (1-\sin^2(x)) \cos(x) dx = \int \cos(x) dx - \int \sin^2(x) \cos(x) dx \\ \int \cos(x) dx &=& \sin(x) + C_1 \end{eqnarray*} Now, we need to integrate $\sin^2(x) \cos(x)$. We do this with the substitution $u = \sin(x)$. \begin{eqnarray*} \int \sin^2(x) \cos(x) dx &=& \int u^2 du = \frac{u^3}{3} + C_1 \\ \int \sin^2(x) \cos(x) dx &=& \frac{\sin^3(x)}{3} + C_1 \\ \int \cos^3(x) dx &=& \sin(x) - \frac{\sin^3(x)}{3} + C \\ \end{eqnarray*}

Two techniques.

(1) 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 $\frac{1}{12}\sin 3x +\frac{3}{4}\sin x$.

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

This reduces to:

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

Letting $u=1+t^2=\sec^2(x/2)$, this is:

$$\int\frac{(2-u)^3}{u^4}\,du$$