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Question

For $n > 0$, Find $$\int_{0}^{2\pi}\frac{x\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$

My Approach

Let $I_{n}$ =$$\int_{0}^{2\pi}\frac{x \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$ I$_{n}$=$\int_{0}^{2\pi}$$\frac{\left(2\pi-x\right)\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$\Longrightarrow$2I$_{n}=$$\int_{0}^{2\pi}$$\frac{2\pi \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$ $\Longrightarrow$I$_{n}=\int_{0}^{2\pi}\frac{\pi \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$

I am stuck here!

Frieren
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Mohan Sharma
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1 Answers1

5

I will assume $n \in \mathbb{N}$. If in the integral $$I_n = \int^{2\pi}_0 \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx,$$ we set $x \mapsto 2\pi - x$ it can be readily seen that $$I_n = \pi \int^{2\pi}_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.$$

Note the integrand $$f(x) = \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x},$$ is both an even and periodic function with a fundamental period of $\pi$. If $f$ is a continuous bounded function with period $\mathfrak{a}$, then $$\int^{b + \mathfrak{a}}_b f(x) \, dx = \int^\mathfrak{a}_0 f(x) \, dx,$$ where $b \in \mathbb{R}$.

Using this result repeatedly on our integral we have \begin{align*} I_n &= \pi \int^{-\pi + 2\pi}_{-\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(periodic with period $2\pi$)}\\ &= \pi \int^\pi_{-\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx\\ &= 2\pi \int^\pi_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(since it is even)}\\ &= 2\pi \int^{-\pi/2 + \pi}_{-\pi/2} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(periodic with period $\pi$)}\\ &= 4 \pi \int^{\pi/2}_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(since it is even)}. \end{align*}

Now setting $x \mapsto \dfrac{\pi}{2} - x$ gives $$I_n = 4 \pi \int^{\pi/2}_0 \frac{\cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.$$ Thus $$I_n + I_n = 4\pi \int^{\pi/2}_0 \frac{\sin^{2n} x + \cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = 4\pi \int^{\pi/2}_{0} \, dx,$$ or $$I_n = \int^{2\pi}_0 \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = \pi^2.$$

omegadot
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  • What would happen if n is real number greater than zero – Mohan Sharma Dec 03 '17 at 05:30
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    When $n$ is a natural number the integrand always remains a continuous bounded function on the interval $[0,2\pi]$. If $n > 0$ where $n$ is real there are many (in fact infinitely many) values for which this will no longer be the case. If, for example, $n = 1/2$, on the interval $[0,2\pi]$ the integrand is unbounded at the points $x = 3\pi/4$ and $7\pi/4$ meaning the definite integral is no longer proper (it becomes improper) and as a result may no longer exist (that is, it diverges). – omegadot Dec 03 '17 at 07:20