Evaluating $\int\frac{\cos^2x}{1+\tan x}\,dx$ I'd like to evaluate the following integral:
$$\int \frac{\cos^2 x}{1+\tan x}dx$$
I tried integration by substitution, but I was not able to proceed. 
 A: Here's another way; rewrite:
$$\int \frac{\cos^2 x}{1+\tan x} \,\mbox{d}x 
= \int \frac{\sec^2 x}{\left(1+\tan x\right) \sec^4 x} \,\mbox{d}x
= \int \frac{\sec^2 x}{\left(1+\tan x\right)\left( 1+\tan^2x\right)^2} \,\mbox{d}x $$
Now set $u = \tan x$ to get:
$$\int \frac{1}{\left(1+u\right)\left( 1+u^2\right)^2} \,\mbox{d}u$$
And you can continue with partial fractions. Tedious, but it works.
A: $$\dfrac{2\cos^2x}{1+\tan x}=\dfrac2{(1+\tan x)(1+\tan^2x)} =\dfrac{1+\tan^2x+1-\tan^2x}{(1+\tan x)(1+\tan^2x)}$$
$$=\dfrac1{1+\tan x}+\dfrac{1-\tan x}{1+\tan^2x}$$
$$=\dfrac{\cos x}{\cos x+\sin x}+\cos^2x(1-\tan x)$$
$\cos^2x(1-\tan x)$ can be managed easily.
Now $\dfrac{2\cos x}{\cos x+\sin x}=1+\dfrac{\cos x-\sin x}{\cos x+\sin x}$
$\dfrac{d(\cos x+\sin x)}{dx}=?$
A: $\displaystyle \int \frac{\cos^2 x}{1+\tan x}dx = \int\frac{\cos^3 x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{(\cos^3 x+\sin ^3 x)+(\cos^3 x-\sin ^3 x)}{\cos x+\sin x}dx$
$\displaystyle = \frac{1}{4}\int (2-\sin 2x)dx+\frac{1}{4}\int\frac{(2+\sin 2x)(\cos x-\sin x)}{\cos x+\sin x}dx$
put $\cos x+\sin x= t$ and $1+\sin 2x = t^2$ in second integral
A: HINT:
$$\frac{\cos^2x}{1+\tan x}=\frac{\cos^3x}{\cos x+\sin x}$$
Now $\cos x+\sin x=\sqrt2\cos\left(x-\dfrac\pi4\right)$
Set $x-\dfrac\pi4=u\implies\cos x=\cos\left(u+\dfrac\pi4\right)=\sqrt2(\cos u-\sin u)$
A: This one always works for rational functions of $\sin x$ and $\cos x$ but can be a bit tedious. Set:
$$ z = \tan x / 2$$
so that
$$ \mathrm{d}x = \frac{2\,\mathrm{d} z}{1 + z^2}$$
$$ \cos x = \frac{1 - z^2}{1 + z^2}$$
$$\sin x = \frac{2z}{1 + z^2}$$
Now, you have a rational fraction in $z$ that you can integrate by standard methods (partial fraction decomposition).
There are often simpler (and trickier) substitutions for this kind of integrals, but this one will always do the job.
