Integrating $\cos^3 (x) \, dx$ I am wondering whether I integrated the following correctly.


*

*$\int \cos^3 x \, dx$
I did 
\begin{align}
\int \cos^3 x \, dx &= \int \cos(x)(1-\sin^{2}x) \, dx \\
&= \int \cos(x)-\sin^{2}x \cos x \, dx \\
&= \sin(x)-\frac{u^{3}}{3} + c, \quad(u=\sin(x)) \\
&= \sin(x)-\dfrac{\sin^{3}x}{3}+c
\end{align}

2.$\int \sin^{3}x \cos^{2}x\,dx$
\begin{align}
\int \sin^{3}x \cos^{2}x\,dx &= \int(1-\cos^2x)(\cos^2x)\sin(x)\,dx \\
&= \int \cos^2x\sin(x)-\cos^4x\sin(x)\,dx, \quad u=\cos(x) \\
&= \dfrac{u^3}{3}-\dfrac{u^5}{5}+c
\end{align}
And plug in my u.

You can even put the formula of $$\cos^{3}x = \frac{\cos3x + 3 \cos x}{4}$$
And then integrate it where you get the answer as
$$ \frac{\sin 3x}{12} + \frac{3 \sin x}{4} + C$$
