Indefinite Integration 1 Question 57

58. For $n>0, \int _0^{2 \pi} \frac{x \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x=$

$(1997,2 M)$

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Solution:

  1. Let $I=\int _0^{2 \pi} \frac{x \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$I=\int _0^{2 \pi} \frac{(2 \pi-x)[\sin (2 \pi-x)]^{2 n}}{[\sin (2 \pi-x)]^{2 n}+[\cos (2 \pi-x)]^{2 n}} d x$

$$ \left[\because \int _0^{a} f(x) d x=\int _0^{a} f(a-x) d x\right] $$

$I=\int _0^{2 \pi} \frac{(2 \pi-x) \cdot \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\Rightarrow \quad I=\int _0^{2 \pi} \frac{2 \pi \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x-\int _0^{2 \pi} \frac{x \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\Rightarrow \quad I=\int _0^{2 \pi} \frac{2 \pi \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x-I$

[from Eq. (i)]

$\Rightarrow \quad I=\int _0^{2 \pi} \frac{\pi \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\Rightarrow \quad I=\pi \int _0^{\pi} \frac{\pi \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$+\int _0^{\pi} \frac{\sin ^{2 n}(2 \pi-x)}{\sin ^{2 n}(2 \pi-x)+\cos ^{2 n}(2 \pi-x)} d x$

$$ \text { using property } $$

$\int _0^{2 a} f(x) d x=\int _0^{a}[f(x)+f(2 a-x)] d x$

$I=\pi \int _0^{\pi} \frac{\sin ^{2 n} x d x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$+\int _0^{\pi} \frac{\sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\Rightarrow \quad I=2 \pi \int _0^{\pi} \frac{\sin ^{2 n} x d x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\Rightarrow \quad I=4 \pi \quad \int _0^{\pi / 2} \frac{\sin ^{2 n} x d x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\Rightarrow \quad I=4 \pi \int _0^{\pi / 2} \frac{\sin ^{2 n}(\pi / 2-x)}{\sin ^{2 n}(\pi / 2-x)+\cos ^{2 n}(\pi / 2-x)} d x$

$\Rightarrow \quad I=4 \pi \int _0^{\pi / 2} \frac{\cos ^{2 n} x}{\cos ^{2 n} x+\sin ^{2 n} x} d x$

On adding Eqs. (ii) and (iii), we get

$$ \begin{array}{rlrl} & & 2 I=4 \pi \int _0^{\pi / 2} \frac{\sin ^{2 n} x+\cos ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \\ \Rightarrow \quad 2 I & =4 \pi \int _0^{\pi / 2} 1 d x=4 \pi[x] _0^{\pi / 2}=4 \pi \cdot \frac{\pi}{2} \\ \Rightarrow \quad & I & =\pi^{2} \end{array} $$



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