# Math Help - How do I solve this $\int_\frac{pi}{2}^0 \frac{cosx}{1 + sin^2x} dx$?

1. ## How do I solve this $\int_\frac{pi}{2}^0 \frac{cosx}{1 + sin^2x} dx$?

How do I solve this $\int_0^\frac{pi}{2} \frac{cosx}{1 + sin^2x} dx$

2. Make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$.

3. Originally Posted by Prove It
Make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$.

$\int \frac{du}{1 + u^2}$
$=tan^{-1}u$
$=tan^{-1}(sin x)\big|_0^{\frac{pi}{2}}$

am I doing this right?

4. Yes, now substitute the terminals...

5. $\int \frac{du}{1 + u^2}$
$=tan^{-1}u$
$=tan^{-1}(sin x)\big|_0^{\frac{pi}{2}}$
$=tan^{-1}(1) - tan^{-1}(0)$
like this?

= pi/4

6. Correct

7. Originally Posted by Prove It
Correct
great success thank you again