IIT-JEE 1996Question 1
For $n>0$, $\displaystyle\int_0^{2\pi}\frac{x\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}\,dx=$ ______.
Definite Integration
Answer$\pi^{2}$
Solution
Concept. King's rule on $[0,2\pi]$ removes $x$. Then $g$ has period $\pi$ and is symmetric about $\frac\pi2$, so $\int_0^{2\pi}g=4\int_0^{\pi/2}g$, and $g(x)+g\big(\frac\pi2-x\big)=1$. Formulas used. (1) $\int_0^{2\pi}x\,g(x)\,dx=\pi\int_0^{2\pi}g(x)\,dx$ if $g(2\pi-x)=g(x)$(2) $\sin^{2n}$, $\cos^{2n}$ are unchanged by $x\to\pi-x$ and $x\to x+\pi$(3) $x\to\frac\pi2-x$ swaps $\sin$ and $\cos$, so $g(x)+g\big(\frac\pi2-x\big)=1$Step 1 - Use (1). With $g=\frac{\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}$, $g(2\pi-x)=g(x)$, so $I=\pi\int_0^{2\pi}g\,dx$. Step 2 - Use (2). $\int_0^{2\pi}g=2\int_0^\pi g=4\int_0^{\pi/2}g$. Step 3 - Use (3). Adding $\int_0^{\pi/2}g$ to itself after $x\to\frac\pi2-x$ gives $2\int_0^{\pi/2}g=\frac\pi2$, so $\int_0^{\pi/2}g=\frac\pi4$. Step 4 - Value. $I=\pi\cdot4\cdot\frac\pi4=\pi^2$. Answer: $\pi^{2}$