How do I solve this [math] \int_\frac{pi}{2}^0 \frac{cosx}{1 + sin^2x} dx[/math]?

**bijosn** · March 26th 2011, 01:28 AM

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

**Prove It** · March 26th 2011, 01:30 AM

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

**bijosn** · March 26th 2011, 02:02 AM

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?

**Prove It** · March 26th 2011, 02:06 AM

Yes, now substitute the terminals...

**bijosn** · March 26th 2011, 02:15 AM

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

= pi/4

**Prove It** · March 26th 2011, 02:27 AM

Correct

**bijosn** · March 26th 2011, 06:05 AM

Originally Posted by Prove It

Correct

great success

thank you again

Thread: How do I solve this [math] \int_\frac{pi}{2}^0 \frac{cosx}{1 + sin^2x} dx[/math]?