# Math Help - Difficult Trigonometric Integral

1. ## Difficult Trigonometric Integral

I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.

$\int {sin^2 x \over \sqrt {1+cosx}}$

Thanks for the help!

2. Originally Posted by Tylerzzz
I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.

$\int {sin^2 x \over \sqrt {1+cosx}}$
Substitute $u = \cos x$. Then $du = -\sin x\,dx$ and the integral becomes

$-\int \frac{\sin x}{\sqrt {1+cosx}}(-\sin x)dx = -\int \frac{\sqrt{1-u^2}}{\sqrt{1+u}}du = -\int\sqrt{1-u}\,du = \frac23(1-u)^{3/2} = \frac23(1-\cos x)^{3/2}$ (plus a constant).

3. Wow! I never really thought about solving it that way. Thanks a lot!

4. EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]

5. Originally Posted by TheCoffeeMachine
EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]
You had me worried for a moment.

6. Hello, Tylerzzz!

Another approach . . .

$\displaystyle \int \frac{\sin^2\!x}{\sqrt{1+\cos x}}\,dx$

We have: . $\dfrac{\sin^2\!x}{\sqrt{1+\cos x}}$

Multiply by $\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}\!:\;\;\dfrac{\sin^2\!x}{\sqrt{1+\cos x}} \cdot\dfrac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{1-\cos^2\!x}}$

. . . $\;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{\sin^2\!x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sin x} \;=\;\sin x\sqrt{1-\cos x}$

The integral becomes: . $\displaystyle \int(1-\cos x)^{\frac{1}{2}}(\sin x\,dx)$

Let: . $u \:=\:1-\cos x \quad\Rightarrow\quad du \:=\:\sin x\,dx$

Substitute: . $\displaystyle \int u^{\frac{1}{2}}\,du \;=\;\tfrac{2}{3}u^{\frac{3}{2}} + C$

Back-substitute: . $\frac{2}{3}(1-\cos x)^{\frac{3}{2}} + C$

7. Originally Posted by Opalg
You had me worried for a moment.
Haha! In my defence, I blame the UCAS for the plight of my eyesight today.