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

• Mar 26th 2011, 01:28 AM
bijosn
How do I solve this $\int_\frac{pi}{2}^0 \frac{cosx}{1 + sin^2x} dx$?
How do I solve this $\displaystyle \int_0^\frac{pi}{2} \frac{cosx}{1 + sin^2x} dx$
• Mar 26th 2011, 01:30 AM
Prove It
Make the substitution $\displaystyle \displaystyle u = \sin{x} \implies du = \cos{x}\,dx$.
• Mar 26th 2011, 02:02 AM
bijosn
Quote:

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

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

am I doing this right?
• Mar 26th 2011, 02:06 AM
Prove It
Yes, now substitute the terminals...
• Mar 26th 2011, 02:15 AM
bijosn
$\displaystyle \int \frac{du}{1 + u^2}$
$\displaystyle =tan^{-1}u$
$\displaystyle =tan^{-1}(sin x)\big|_0^{\frac{pi}{2}}$
$\displaystyle =tan^{-1}(1) - tan^{-1}(0)$
like this?

= pi/4
• Mar 26th 2011, 02:27 AM
Prove It
Correct :)
• Mar 26th 2011, 06:05 AM
bijosn
Quote:

Originally Posted by Prove It
Correct :)

great success(Nod) thank you again