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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.

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    The "textbook" substitution here is $z = \tan x/2$. Everytime you have a rational function of $\cos x$ and $\sin x$ you can perform this substitution There are other (often simpler) tricks, like the rule of Bioche, etc but this one always works. – 2017-01-03
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    Obligatory comment: You do not solve an integral, you evaluate it. – 2017-01-03

5 Answers 5

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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.

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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.

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$$\dfrac{2\cos^2x}{1+\tan x}=\dfrac2{(1+\tan x)(1+\tan^2x)} =\dfrac{1+\tan^2x+1-\tan^2x}{(1+\tan x)(1+\tan^2x)}$$

$$=\dfrac1{1+\tan x}+\dfrac{1-\tan x}{1+\tan^2x}$$

$$=\dfrac{\cos x}{\cos x+\sin x}+\cos^2x(1-\tan x)$$

$\cos^2x(1-\tan x)$ can be managed easily.

Now $\dfrac{2\cos x}{\cos x+\sin x}=1+\dfrac{\cos x-\sin x}{\cos x+\sin x}$

$\dfrac{d(\cos x+\sin x)}{dx}=?$

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$\displaystyle \int \frac{\cos^2 x}{1+\tan x}dx = \int\frac{\cos^3 x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{(\cos^3 x+\sin ^3 x)+(\cos^3 x-\sin ^3 x)}{\cos x+\sin x}dx$

$\displaystyle = \frac{1}{4}\int (2-\sin 2x)dx+\frac{1}{4}\int\frac{(2+\sin 2x)(\cos x-\sin x)}{\cos x+\sin x}dx$

put $\cos x+\sin x= t$ and $1+\sin 2x = t^2$ in second integral

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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 $x-\dfrac\pi4=u\implies\cos x=\cos\left(u+\dfrac\pi4\right)=\sqrt2(\cos u-\sin u)$

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    This is not a good hint as OP don't know how to proceed with the "$\cos^3x$"... – 2017-01-03
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    @DeepSea, Can he/she use $$\cos^3x=2\sqrt2(\cos u-\sin u)^3$$ – 2017-01-03