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?
 A: HINT:
$$\int x\dfrac{\cos x}{\sin x(1+\sin^2x)}dx$$
$$=x\int\dfrac{\cos x}{\sin x(1+\sin^2x)}dx-\int\left(\dfrac{dx}{dx}\cdot\int\dfrac{\cos x}{\sin x(1+\sin^2x)}dx\right)dx$$
Set $\sin x=u$ 
For $\dfrac1{u(u^2+1)}=\dfrac{u^2+1-u^2}{u(u^2+1)}=?$
A: First Start with Basic Integration by Parts:
$$ \int_0^{\frac{\pi}{2}}\dfrac{x \cos x}{\sin{x}+\sin^3 x}\, dx = -\frac{\pi}{4}\log 2 - \int_0^{\frac{\pi}{2}} \left( \log{\sin{x}} - \frac{\log(1+\sin^2 x)}{2}\, dx \right) $$
$$$$

Since,
  $$\int_0^{\frac{\pi}{2}} \log\sin x\, dx = -\frac{\pi}{2} \log 2 $$
$$\therefore\int_0^{\frac{\pi}{2}}\frac{x \cos x}{\sin x+\sin^3 x} \, dx = \frac{\pi}{4} \log 2 + \int_0^{\frac{\pi}{2}} \frac{\log(1+\sin^2 x)}{2}\, dx \tag 1$$

$$$$

Now, let (for some $k$):
  $$I(k) = \int_0^{\frac{\pi}{2}} \frac{\log(1+\sin^2 x + k \sin^2 x)}{2}\, dx \tag 2$$

In order to find $I(k)$:
$$\frac{\partial I}{\partial k} = \int_0^{\frac{\pi}{2}} \frac{\sin^{2}{x}\, dx}{1+\sin^2 x +k\, \sin^2 x} $$
$$\implies I'(k) = \int_0^{\frac{\pi}{2}} \frac{1}{k+1}\left(1-\frac{\sec^2 x}{1+(k+2)\tan^2 x}\right)\, \, dx $$
$$=\frac{x}{k+1}-\frac{\tan^{-1}{(\sqrt{k+2}\,\tan x)}}{(k+1)\, \sqrt{k+2}}\bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{2k+2}\left(1-\frac{1}{\sqrt{k+2}}\right)$$

Now just integrate (using Calculus 1 knowledge):
  $$I(k)= \frac{\pi}{2}\left(\log{(k+1)}-\log\left(\frac{\sqrt{k+2}-1}{\sqrt{k+2}+1}\right)\right)+C=\frac{\pi}{2}\, \log\left(2\, (\sqrt{k+2}+1)^2\right)+C$$
$$\therefore I(k) = \frac{\pi}{2}\, \log{\left(2 (\sqrt{k+2}+1)^2\right)}+C \tag 3$$

From this step forward, we will solve for $C$:
To find $C$, substitute $k=-1$. 
From $(2)$, we have:
$$\therefore I(-1) = 0$$
From $(3)$, we have: 
$$I(-1) = \frac{3\,\pi}{2}\log 2+C$$
$$\implies C= - \frac{3\,\pi}{2}\log 2$$
$$\therefore I(0) = \int_0^{\frac{\pi}{2}} \log(1+\sin^{2}{x})\, dx = \frac{\pi}{2}\, \log\left(2\, (\sqrt{2}+1)^2\right) - \frac{3\,\pi}{2} \log 2$$
Now use $(1)$:
$$\therefore\int_0^{\frac{\pi}{2}}\frac{x \cos x\, dx}{\sin x +\sin^3 x} = \frac{\pi}{4}\,\log 2 + \frac{\pi}{4}\, \log\left(2\, (\sqrt{2}+1)^2\right) - \frac{3\,\pi}{4}\log 2$$
$$= \frac{\pi}{4}\log\left(\frac{3+2\,\sqrt{2}}{2}\right)$$
A: 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
$$\int_0^{\pi/2} d\theta \frac{\cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \frac{\pi}{2} \log{(1+\sqrt{2})} - \frac{\pi}{4} \log{2} = \frac{\pi}{4} \log{\left (\frac{3}{2} + \sqrt{2} \right )}$$
A: 
In the spirit of both (i) the solid answer posted by @yagnapatel and (ii) THIS ANSWER, we proceed here by using the technique of Differentiating Under the Integral Sign to evaluate the integral of interest.  


STEP 1:
Let the integral of interest $I$ be given by
$$I=\int_0^{\pi/2} \frac{\theta \cos \theta}{\sin \theta +\sin^3 
\theta}\,d\theta \tag 1$$
First, integrating $(1)$ by parts with $u=\theta$ and $v=\log (\sin \theta)-\frac12 \log (1+\sin^2 \theta)$ yields
$$\begin{align}
I&=-\frac{\pi}{4}\log 2+\int_0^{\pi/2}\left(\frac12 \log (1+\sin^2 \theta)-\log (\sin \theta)\right)\,d\theta\\\\
&=\frac{\pi}{4}\log 2+\frac12\int_0^{\pi/2} \log (1+\sin^2 \theta)\,d\theta \tag 2
\end{align}$$

STEP 2:
Second, we examine the integral $J(a)$ that is defined as 
$$J(a)=\int_0^{\pi/2}\log (a+\sin^2 \theta)\,d\theta$$
and note that $J(1)$ is the integral that appears on the right-hand side of $(2)$.  
In addition, we observe that $J(0)=-\pi\log 2$, and $J'(a)$ is given by
$$J'(a)=\int_0^{\pi/2} \frac{1}{a+\sin^2\theta}\,d\theta$$
Now, in THIS ANSWER, I showed that $J'(a)$ is given by
$$\begin{align}
J'(a)&=\int_0^{\pi/2}\frac{1}{a+\sin^2\theta}\,d\theta\\\\
&=\text{sgn}(a)\frac{\pi}{2\sqrt{a(a+1)}} \tag 3
\end{align}$$

Note that this result could also be obtained using the well-known Weierstrass Substitution.


STEP 3:
Third, we integrate $(3)$ and make use of $J(0)=-\pi \log 2$ to find that for $a>0$
$$J(a)=\pi \log(\sqrt{a}+\sqrt{1+a})-\pi \log 2$$
and therefore 
$$J(1)=\pi\log(1+\sqrt{2})-\pi \log 2 \tag 4$$

STEP 4:
Finally, we substitute $(4)$ into $(2)$ to reveal
$$\bbox[5px,border:2px solid #C0A000]{I=\frac{\pi}{2}\log\left(1+\frac{\sqrt{2}}{2}\right)}$$
which agrees with the result reported by others since $\left(1+\frac{\sqrt{2}}{2}\right)^2=\frac32 +\sqrt{2}$!   And we're done.
A: $$
\int \frac {\theta \cos \theta  } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \underbrace{\int \theta \, dx = \theta x - \int x\,d\theta}_{\text{integration by parts}}
$$
$$
dx = \frac { \cos \theta  } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \frac{du}{u+u^3} = \frac{du}{u(1+u^2)} = \left( \frac A u + \frac{Bu+C}{1+u^2} \right)\,du
$$
$$
\frac{Bu}{1+u^2} \, du = \frac B 2\cdot \frac{dw} w
$$
$$
\int \frac C {1+u^2} \, du = C \arctan u +\text{constant}
$$
You need to do a bit of algebra to find the three coefficients $A,B,C$, and then put everything together.
I'm getting $A=1$, $B=-1$, $C=0$, so
$$
\int \left( \frac 1 u - \frac u {1+u^2}\right)\, du = \log |u| - \frac 1 2 \log(1+u^2) + \text{constant}
$$
but we don't need the constant: in integration by parts we need only one antiderivative, not all of them.
This is
$$
\log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta).
$$
So the integral is
$$
\theta\left( \log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta) \right) - \int \left( \log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta) \right) \, d\theta + \text{constant}
$$
I haven't taken it beyond that yet, and I don't know whether it can be done is closed form.
However, sometimes it is possible to find a definite integral even if you can't find the indefinite integral.
