How to compute $\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}dt$?

Question

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?

Answer 1

The substitution $y=\frac{\pi}{2}-t$ solves it... If you do this substitution, you get:

$$\int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}dt= \int_0^{\pi/2}\frac{\cos^n y}{\cos^n y+\sin^n y}dy \,.$$

Answer 2

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 $n$.

Answer 3

In general

Dividing both the numerator and denominator by $\cos ^n x$ converts $$ \begin{aligned} I_{n} =&\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{n} t} d t \\ \begin{aligned} \\ \end{aligned} \\\stackrel{t\mapsto\frac{\pi}{2}-t}{=} &\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot ^{n} t} d t \\ =&\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{n} t}{\tan ^{n} t+1} d t \\ 2 I_{n} =&\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{1+\tan ^{n} t}+\frac{\tan ^{n} t}{\tan ^{n} t+1}\right) d t = \int_{0}^{\frac{\pi}{2}} 1 d t \\ I_{n} =&\frac{\pi}{4} \end{aligned} $$