Solve the integral $\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$ Question
Solve the integral,$$\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$$
Attempt
I converted the equation in terms of $\sin(x)$ and $\cos(x)$ using the definition of $\tan(x)$ and $\cot(x)$, and then applied the substitution $t=\sin(x)$, however, this has proved itself to be quite a difficult integral to resolve in and of itself.
I would be great full for any suggestions of a more compact method. Any hints would be also be greatly appreciated.
 A: Integrate as follows
\begin{align}
&\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx\\
=& \int_{0}^{\frac{\pi}2} \frac{\sin^3x+ \cos^3x}{\tan^3x+\cot^3x} dx
= \int_{0}^{\frac{\pi}2} \frac{(\sin x+ \cos x)(1-\sin x\cos x)}{(\tan x+ \cot x)(\tan^2x+\cot^2x-1)} dx\\
=& \int_{0}^{\frac{\pi}2} \frac{(\sin x+ \cos x)(1-\sin x\cos x)(\sin x\cos x)^3}{1-3\sin^2x\cos^2x} dx\\
 =& \ \frac14\int_{0}^{\frac{\pi}2} \frac{(\sin x+ \cos x)(2-\sin 2x)\sin^32x}{4-3\sin^22x} dx\>\>\>\>\>\>\>t= \sin x-\cos x\\
=& \ \frac14\int_{-1}^1 \frac{(1+t^2)(1-t^2)^3}{4-3(1-t^2)^2} dt
= \int_{-1}^1 \left( \frac{t^4}{12}+\frac1{36}+\frac{3t^2-1}{9(3t^4-6t^2-1)} \right)dt\\
=& \ \frac4{45}+\frac{2^{3/2}}{3^{9/4} }
\left( \tan^{-1}\sqrt{2\sqrt3+3}- \tanh^{-1}\sqrt{2\sqrt3-3}\right)
\end{align}
A: Clear
$$ \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx=2\int_{0}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx. $$
Let
$$ I=\int_{0}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x}dx. \tag1$$
Then
$$ I=\int_{0}^{\frac{\pi}2} \frac{\cos^3x}{\tan^3x+\cot^3x}dx. \tag2$$
Adding (1) to (2) gives
\begin{eqnarray}
2I&=&\int_{0}^{\frac{\pi}2} \frac{\sin^3x+\cos^3x}{\tan^3x+\cot^3x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{(\sin x+\cos x)(1-\sin x\cos x)}{(\tan x+\cot x)(\tan^2x+\cot^2x-1)} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^3 x\cos^3 x(\sin x+\cos x)(1-\sin x\cos x)}{\sin^4x+\cos^4x-\sin^2x\cos^2x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^3 x\cos^3 x(\sin x+\cos x)(1-\sin x\cos x)}{1-3\sin^2x\cos^2x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^3 x}{1-3\sin^2x\cos^2x} dx+\int_{0}^{\frac{\pi}2} \frac{\sin^3 x\cos^4 x}{1-3\sin^2x\cos^2x} dx-\int_{0}^{\frac{\pi}2} \frac{\sin^5 x\cos^4 x}{1-3\sin^2x\cos^2x} dx-\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^5 x(1-\sin x\cos x)}{1-3\sin^2x\cos^2x} dx\\
&=&2\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^3 x}{1-3\sin^2x\cos^2x} dx-2\int_{0}^{\frac{\pi}2} \frac{\sin^5 x\cos^4 x}{1-3\sin^2x\cos^2x} dx\\
&=:&2J_1-2J_2.
\end{eqnarray}
Note that, under $t=\sin x$,
\begin{eqnarray}
J_1&=&\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^3 x}{1-3\sin^2x\cos^2x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^2 x}{1-3\sin^2x\cos^2x} d\sin x\\
&=&\int_{0}^{1} \frac{t^4(1-t^2)}{1-3t^2(1-t^2)} dt
\end{eqnarray}
which is not hard to handle by partial fractions. You can do the same thing for $J_2$. I omit the details.
