Calculating the integral of a trigonometric function I have a problem calculating the following integral:

$$\int \cfrac{\text dx}{(\sin x+ \cos x)^3}$$

Can somebody help me, please? Thank you in advance!
 A: Note that $\sin(x)+\cos(x) = \sqrt{2} \sin(x+\pi/4)$, this reduces the problem to the problem of integrating $1/(\sin x)^3$, now let $u=cos x$ we find that the integral reduces to the integral of $1/(1-u^2)^2$.
A: $$\sin x+\cos x=\sqrt2(\frac{\sqrt2}2\sin x+\frac{\sqrt2}2\cos x)=\sqrt2\sin(x+\frac\pi4)$$
$$\int\frac{dx}{2\sqrt2\sin^3(x+\frac\pi4)}=\frac{\sqrt2}4\int\csc^3(x+\frac\pi4)dx$$
Integration by parts should take care of it from there.
A: As mentioned, $\sin(x)+\cos(x)=\sqrt{2}\sin(x+\pi/4)$. Therefore,
$$
\begin{align}
\int\frac{\mathrm{d}x}{(\sin(x)+\cos(x))^3}
&=\frac1{2\sqrt{2}}\int\frac{\mathrm{d}x}{\sin^3(x+\pi/4)}\\
&=-\frac1{2\sqrt{2}}\int\frac{\mathrm{d}\cos(x+\pi/4)}{\sin^4(x+\pi/4)}\\
&=-\frac1{2\sqrt{2}}\int\frac{\mathrm{d}u}{(1-u^2)^2}\\
&=-\frac1{8\sqrt{2}}\int\left(\frac1{1-u}+\frac1{1+u}+\frac1{(1-u)^2}+\frac1{(1+u)^2}\right)\mathrm{d}u\\
&=\frac1{8\sqrt{2}}\left(\log\left(\frac{1-u}{1+u}\right)+\frac1{1+u}-\frac1{1-u}\right)+C\\
&=\frac1{8\sqrt{2}}\left(\log\left(\frac{1-u^2}{(1+u)^2}\right)-\frac{2u}{1-u^2}\right)+C\\
&=\frac1{4\sqrt{2}}\left(\log\left(\tan(x/2+\pi/8)\right)-\cot(x+\pi/4)\csc(x+\pi/4)\right)+C
\end{align}
$$
