<s>[INST] Find $\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ My Calc 2 teacher wasn't able to solve this:
$$\int_0^\frac{\pi}{2} \frac {\theta \cos \theta  } { \sin \theta + \sin ^ 3 \theta }\:d\theta$$
Can someone help me solve this?
 [/INST] You can transform the integral by parts to get
$$\int_0^{\pi/2} d\theta \frac{\theta \cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \int_0^{\pi/2} d(\sin{\theta}) \left (\frac1{\sin{\theta}}-\frac{\sin{\theta}}{1+\sin^2{\theta}} \right )\theta \\ = -\frac{\pi}{4} \log{2} - \int_0^{\pi/2} d\theta \left [\log{(\sin{\theta})} -  \frac12 \log{(1+\sin^2{\theta})} \right ]$$
Now,
$$\int_0^{\pi/2} d\theta \log{(\sin{\theta})} = - \frac{\pi}{2} \log{2}$$
$$\int_0^{\pi/2} d\theta \,\log{(1+\sin^2{\theta})} = \frac{\pi}{2} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k 2^{2 k}} \binom{2 k}{k} $$
Defining
$$f(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k 2^{2 k}} \binom{2 k}{k} x^k$$
we find that
$$f'(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2^{2 k}} \binom{2 k}{k} x^{k-1} = \frac1{x} \left (\frac1{\sqrt{1+x}}-1 \right )$$
so that
$$f(x) = \log{\left [\frac{\sqrt{1+x}-1}{ x \left (\sqrt{1+x}+1\right )} \right ]}+C$$
$$\lim_{x \to 0} f(x) = 0 \implies C=\log{4} $$
Thus, putting this all together, we get
$$\int_0^{\pi/2} d\theta \frac{\cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \frac{\pi}{2} \log{(1+\sqrt{2})} - \frac{\pi}{4} \log{2} = \frac{\pi}{4} \log{\left (\frac{3}{2} + \sqrt{2} \right )}$$
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