Solving Trig Integral with Square Root

Question

The problem is:

$$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sin^2(x) }{\sqrt{1-\cos(x)}}~dx$$

I'm lost. I tried using the half angle formula for sin but eventually got confused along the way.

Answer 1

Many tricks we could apply here. If we know that $1-\cos^2(x)=\sin^2(x)$ as the pythagorean theorem, then

$$\begin{align}\frac{\sin^2(x)}{\sqrt{1-\cos(x)}}&=\frac{\sin^2(x)\sqrt{1+\cos(x)}}{\sqrt{1-\cos(x)}\sqrt{1+\cos(x)}}\\&=\frac{\sin^2(x)\sqrt{1+\cos(x)}}{\sqrt{1-\cos^2(x)}}\\&=\frac{\sin^2(x)\sqrt{1+\cos(x)}}{|\sin(x)|}\\&=|\sin(x)|\sqrt{1+\cos(x)}\end{align}$$

And depending on the interval you are integrating over, $|\sin(x)|=\pm\sin(x)$, then apply the substitution $u=\cos(x)$.

Answer 2

Another trick. Note that: $$ \frac{\sin^2x}{\sqrt{1-\cos x}}=\frac{2\cos^2 \frac{x}{2}\sin^2\frac{x}{2} }{\sqrt{1-1+2\sin^2\frac{x}{2}}}=2\sqrt{2}\cos^2\frac{x}{2}\sin\frac{x}{2} $$ Than use the substitution $$ u=\cos \frac{x}{2} \quad \rightarrow \quad du=\frac{1}{2}\sin\frac{x}{2} dx $$

Answer 3

Substitution: $t = tan(x/2)$

$\frac{\mathrm{d} t}{\mathrm{d} x} = \frac{1}{2}sec^2\frac{x}{2}$

$ sinx = \frac{2t}{1+t^2}$

$ cosx = \frac{1-t^2}{1+t^2}$

Then we have:

$ I = \int_{1/\sqrt{3}}^{1}(\frac{2t}{1+t^2})^2*\frac{1}{\sqrt{1- \frac{1-t^2}{1+t^2}}}*\frac{2}{1+t^2}dt$

This simplifies to: $$\frac{8}{\sqrt{2}}\int_{\frac{1}{\sqrt{3}}}^{1}\frac{t}{(1+t^2)^\frac{5}{2}}dt$$

Which we can integrate to: $$I = \frac{-8}{3\sqrt{2}}\left [ \frac{1}{(1+t^2)^\frac{3}{2}} \right ]_{\frac{1}{\sqrt{3}}}^{1}$$

This evaluates to: $I = \sqrt{\frac{3}{2}} - \frac{2}{3}$