Integrate $\int \frac{\cos^2x}{1+\tan x}\,dx$

Question

I tried getting $\cos^2x$ in terms of $\sec^2x$ and tried substituting, as the derivative of $\tan x$ is $\sec^2x$. But it got way too complicated and long.

So, what would be a better approach for it?

Answer 1

hint

$$\cos^2 (x)=\frac {1}{1+\tan^2 (x)} $$

put $t=\tan (x) $to get $$\int \frac {dt}{(1+t)(1+t^2)^2} $$

use partial decomposition and by parts integration for $$\int \frac {dt}{(1+t^2)^2} =\int \frac {1+t^2-t^2}{(1+t^2)^2}dt$$

Answer 2

HINT: your Integrand is equal to $$ \frac { \left( 1- \left( \tan \left( \frac x 2 \right) \right)^2 \right)^2}{ \left( 1+ \left( \tan \left( \frac x 2 \right) \right)^2 \right)^2 (1+\tan(x)) } $$ also you must convert $\tan$ into $$\tan\left( \frac x 2 \right)$$

Answer 3

Here's an interesting substitution. It's not the most elegant method, but the resulting integrals should be pretty straightforward.

$$\frac{\cos^2(x)}{1+\tan(x)}$$

$$=\frac{\cos^3(x)}{\sin(x)+\cos(x)}$$

Noting that $\sin(x+\frac\pi 4)=\frac{\sqrt 2}2\sin(x)+\frac{\sqrt 2}2\cos(x)$ by the angle sum identity.

$$=\frac{\cos^3(x)}{\sqrt 2\sin(x+\frac\pi 4)}$$

We can now substitute $u=x+\frac\pi 4$.

$$=\frac{\cos^3(u-\frac\pi 4)}{\sqrt 2\sin(u)}$$

Using another angle sum expansion $\cos(u-\frac\pi 4)=\frac{\sqrt 2}2\cos(u)+\frac{\sqrt 2}2\sin(u)$,

$$=\frac{(\cos(u)+\sin(u))^3}{4\sin(u)}$$

$$=\frac 14\left[\frac{\cos^3(u)}{\sin(u)}+3\cos^2(u)+3\cos(u)\sin(u)+\sin^2(u)\right]$$

The first term permits a substitution for $\sin u$, and the rest should not be any trouble to integrate.

Answer 4

If all else fails, the substitution suggested in the answer by Dr. Sonnhard Graubner will do the job, but he gives a hint that is perhaps really more of a sketch rather than a hint, and may be comprehensible only to someone who already knows that substitution.

This was introduced by Euler in the 1700s and is sometimes mistakenly reported in textbooks to have been introduced by Weierstrass in the 1800s.

It will work if all else fails, but I haven't ruled out the possibility that something simpler will work.

It is a tangent half-angle substitution: \begin{align} \tan \frac x 2 & = u \\[10pt] & \text{From this, it follows by trigonometric identities that} \\[10pt] \sin x & = \frac{2u}{1+u^2} \\[10pt] \cos x & = \frac{1-u^2}{1+u^2} \\[10pt] \text{and so } \tan x & = \frac{2u}{1-u^2} \\[10pt] \text{and, crucially, } dx & = \frac{2\,du}{1+u^2}. \end{align}

After simplifying all the fractions you'll have $$ \int \frac{\cdots\cdots\cdots}{1-u^2+2u} \, du = \int \frac{\cdots\cdots\cdots}{(1+\sqrt 2 - u)(1-\sqrt 2 - u)} \, du. $$ However, if I'm not mistaken, you get a 4th-degree polynomial in $u$ in the numerator, so before going to partial fractions, do long division to get $$ (\text{2nd-degree polynomial in }u) + \frac{\text{remainder}}{(1+\sqrt 2 - u)(1-\sqrt 2 - u)} $$ and then use partial fractions.

After integrating, you will need to convert back to a function of $x,$ and simplifying after that may require a bunch of routine trigonometric identities. (If you have a definite integral, then you won't need to do that last part because you will have made the appropriate changes in the bounds of integration before you get there.)