andu1854
  • andu1854
Evaluate the following integral I will draw it out, but its set up like this: sin^2 x dx ---------- sqrt(1-cosx) I need help setting up what u will equal and going from there...
Mathematics
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andu1854
  • andu1854
Evaluate the following integral I will draw it out, but its set up like this: sin^2 x dx ---------- sqrt(1-cosx) I need help setting up what u will equal and going from there...
Mathematics
jamiebookeater
  • jamiebookeater
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andu1854
  • andu1854
|dw:1434089857664:dw|
andu1854
  • andu1854
From here, I was able to conjugate the denominator to get: |dw:1434089965873:dw|
ganeshie8
  • ganeshie8
i think below identity might simplify the integrand \[1-\cos x ~=~ 2\sin^2(\frac{x}{2})\]

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ganeshie8
  • ganeshie8
also this \[\sin x ~~=~~2\sin(\frac{x}{2})\cos(\frac{x}{2}) \]
ganeshie8
  • ganeshie8
play wid them and see if you can get a nice looking expression for the integrand
anonymous
  • anonymous
@andu1854 I think you meant to multiply by the conjugate \(\sqrt{1+\cos x}\)? So you have \[\int_{\pi/3}^{\pi/2}\frac{\sin^2x}{\sqrt{1-\cos x}}\,dx=\int_{\pi/3}^{\pi/2}\frac{\sin^2x\sqrt{1+\cos x}}{\sqrt{1-\cos^2 x}}\,dx\] The denominator simplifies nicely: \[\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|=\sin x\] where the last equality holds because \(\sin x>0\) for \(\dfrac{\pi}{3}\le x\le\dfrac{\pi}{2}\). The integral simplifies further: \[\int_{\pi/3}^{\pi/2}\frac{\sin^2x\sqrt{1+\cos x}}{\sin x}\,dx=\int_{\pi/3}^{\pi/2}\sin x\sqrt{1+\cos x}\,dx\] and you can deal with this using a simple substitution.

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