Definite Integration Question 57
Question 57
- 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:
- 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
$$ \begin{array}{rlrl} & & 2 I=4 \pi \int_{0}^{\pi / 2} \frac{\sin ^{2 n} x+\cos ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \ \Rightarrow \quad 2 I & =4 \pi \int_{0}^{\pi / 2} 1 d x=4 \pi[x]_{0}^{\pi / 2}=4 \pi \cdot \frac{\pi}{2} \ \Rightarrow \quad & I & =\pi^{2} \end{array} $$
