andu1854 one year ago 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...

1. andu1854

|dw:1434089857664:dw|

2. andu1854

From here, I was able to conjugate the denominator to get: |dw:1434089965873:dw|

3. ganeshie8

i think below identity might simplify the integrand $1-\cos x ~=~ 2\sin^2(\frac{x}{2})$

4. ganeshie8

also this $\sin x ~~=~~2\sin(\frac{x}{2})\cos(\frac{x}{2})$

5. ganeshie8

play wid them and see if you can get a nice looking expression for the integrand

6. 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.