# Math Help - [SOLVED] Evalute integral

1. ## [SOLVED] Evalute integral

Hi!

Problem: $\int_{0}^{\frac{\pi}{4}} \frac{dx}{1+sin(x)cos(x)} \; dx$

Just point me in the right direction.

Thx!

2. $\int_{0}^{\frac{\pi}{4}}\frac{1}{1+sin(x)cos(x)}dx$

Let $x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du$

$\int\frac{1}{u^{2}+u+1}du$

Now, let $u=\frac{2w-1}{2}, \;\ du=dw$

Getting:

$4\int\frac{1}{3+4w^{2}}dw$

Now, it should be easier. Don't forget to change your limits of integration.

3. Hi!

I dont follow from second to third line.

4. Originally Posted by Twig
Hi!

I dont follow from second to third line.
Because $\sin(x)\cos(x)=\frac{\sin(2x)}{2}$

If you can't see it, first substitute $t=2x$
And then apply Weierstrass substitution ( $w=\tan\tfrac t2$)
It's equivalent.

5. Originally Posted by Twig
Hi!

I dont follow from second to third line.
$\sin(x)\cos(x) = \frac{\tan x}{1+\tan^{2}x} = \frac {u}{1+u^{2}}$

so $\int \frac {dx}{1+ \cos x \sin x} = \int \frac {\frac{du}{1+u^{2}}}{1 + \frac {u}{1+u^{2}}}$ $= \int \frac {du}{1+u+u^{2}}$

6. $\int \frac{dx}{ 1 + \sin{x} \cos{x}}$

$= \int \frac{2dx}{2 + \sin2x }$

Sub $x = \frac{\pi}{4} - u$

$- \int \frac{2du}{ \sin^2{u} + \cos^2{u} + 2\cos^2{u}}$

$= -\int \frac{2 \sec^2{u} du}{3 + \tan^2{u}}$

$= \frac{-2 }{ \sqrt{3}} \tan^{-1}{\frac{tan(\pi/4 - x)}{\sqrt{3}}}$