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Calculate the Integral for all n=1,...n

$\int _{ 0 }^{ \pi/2 }{ \frac { \sin^n(x) }{ \sin^n(x)+ \cos^n(x) } } dx$

hint: substituion with $x= (\pi/2)-t$

I tried it many times, but always ended up with the Integralsinus $Si(x)$ which i cant use. Thanks for help

Dando18
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johnka
  • 569

2 Answers2

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Let $x=\frac{\pi}{2}-t$. Then

\begin{align} \int_0^\frac{\pi}{2}\frac{\sin^nx}{\sin^nx+\cos^nx}dx&=-\int_{\frac{\pi}{2}}^0\frac{\sin^n\left(\frac{\pi}{2}-t\right)}{\sin^n\left(\frac{\pi}{2}-t\right)+\cos^n\left(\frac{\pi}{2}-t\right)}dt\\ &=\int_0^\frac{\pi}{2}\frac{\cos^nt}{\cos^nt+\sin^nt}dt\\ &=\int_0^\frac{\pi}{2}\frac{\cos^nx}{\sin^nx+\cos^nx}dx \end{align}

Therefore,

\begin{align} \int_0^\frac{\pi}{2}\frac{\sin^nx}{\sin^nx+\cos^nx}dx&=\frac{1}{2}\int_0^\frac{\pi}{2}\frac{\sin^nx}{\sin^nx+\cos^nx}dx+\frac{1}{2}\int_0^\frac{\pi}{2}\frac{\cos^nx}{\sin^nx+\cos^nx}dx\\ &=\frac{1}{2}\int_0^\frac{\pi}{2}dx\\ &=\frac{\pi}{4} \end{align}

CY Aries
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Let

$$I(n)=\int _{ 0 }^{ \frac{\pi}2 }{ \frac { \sin^n(x) }{ \sin^n(x)+\cos^n(x) } } dx$$

Using the substitution $x\to\pi/2-x$, we find

$$I(n)= \int _{ 0 }^{ \frac{\pi}2 }{ \frac { \cos^ n(x) }{ \sin^n(x)+\cos^n(x) } }dx$$

What happens when you add these two forms of $I(n)$?

Argon
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