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

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March 26th 2011, 01:28 AM
bijosn

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

How do I solve this $\int_0^\frac{pi}{2} \frac{cosx}{1 + sin^2x} dx$
March 26th 2011, 01:30 AM
Prove It

Make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$ .
March 26th 2011, 02:02 AM
bijosn

Quote:

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?
March 26th 2011, 02:06 AM
Prove It

Yes, now substitute the terminals...
March 26th 2011, 02:15 AM
bijosn

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

= pi/4
March 26th 2011, 02:27 AM
Prove It

Correct :)
March 26th 2011, 06:05 AM
bijosn

Quote:

Originally Posted by Prove It

Correct :)

great success(Nod) thank you again

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