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## Homework Statement

∫[itex]\frac{sin\Theta}{1+sin^2\Theta}[/itex] d[itex]\Theta[/itex] on [0,[itex]\frac{\pi}{2}[/itex]]

## Homework Equations

I substituted:

u=cos[itex]\Theta[/itex]

du=-sin[itex]\Theta[/itex]

## The Attempt at a Solution

= ∫ [itex]\frac{sin\Theta}{1+1-cos^2\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-cos\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-u^2}[/itex]

= [itex]\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}[/itex] <--- I'm pretty sure I shouldn't have done this.

And that's as far as I've gotten.

Any tips greatly appreciated.