How to compute $\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}dt$? Calculating with Mathematica, one can have
$$\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}\,\mathrm dt=\frac{\pi}{4}.$$


*

*How can I get this formula by hand? Is there any simpler idea than using $u = \sin t$?    

*Is there a simple way to calculate
$$
\int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}\,\mathrm dt
$$
for $n>3$?

*Could anyone come up with a reference for this exercise?  

 A: The substitution $y=\frac{\pi}{2}-t$ solves it... If you do this substitution, you get:
$$\int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}dt= \int_0^{\pi/2}\frac{\cos^n y}{\cos^n y+\sin^n y}dy \,.$$
A: Use the Calculus identity that $$f(x)=f(a-x),$$ and let  $$I=\int_0^\frac{\pi}{2} \frac{\sin^3t}{\sin^3t+\cos^3t}dt.$$ Then, $$f(t)=f(\frac{\pi}{2}-t)=\frac{\sin^3(\frac{\pi}{2}-t)}{\sin^3(\frac{\pi}{2}-t)+\cos^3(\frac{\pi}{2}-t)}=\frac{\cos^3t}{\cos^3t+\sin^3t}$$Thus, $$I=\int_0^\frac{\pi}{2} \frac{\cos^3t}{\cos^3t+\sin^3t}dt.$$ So we have $$2I=\int_0^\frac{\pi}{2} \frac{\sin^3t}{\sin^3t+\cos^3t}dt+\int_0^\frac{\pi}{2} \frac{\cos^3t}{\cos^3t+\sin^3t}dt=\int_0^\frac{\pi}{2}dt=\frac{\pi}{2}.$$ So $$I=\frac{\pi}{4}.$$ Note that this is true for any natural number $n$.
A: Dividing both the numerator and denominator by $\cos ^n x$  converts
$$
\begin{aligned}
I_{n} =&\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{n} t} d t \\
\begin{aligned}
\\
\end{aligned} \\\stackrel{t\mapsto\frac{\pi}{2}-t}{=} &\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot ^{n} t} d t \\
=&\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{n} t}{\tan ^{n} t+1} d t \\
2 I_{n} =&\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{1+\tan ^{n} t}+\frac{\tan ^{n} t}{\tan ^{n} t+1}\right) d t \\
I_{n} =&\frac{\pi}{4}
\end{aligned}
$$
