3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Alexandre C. Jan 3 '17 at 10:32
  • $\begingroup$ Obligatory comment: You do not solve an integral, you evaluate it. $\endgroup$ – s.harp Jan 3 '17 at 11:10
2
$\begingroup$

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

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

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

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

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

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ This is not a good hint as OP don't know how to proceed with the "$\cos^3x$"... $\endgroup$ – DeepSea Jan 3 '17 at 8:49
  • $\begingroup$ @DeepSea, Can he/she use $$\cos^3x=2\sqrt2(\cos u-\sin u)^3$$ $\endgroup$ – lab bhattacharjee Jan 3 '17 at 8:50
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.