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JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1
For n>0n > 0,∫02πxsin⁡2nxsin⁡2nx+cos⁡2nxdx=\int\limits_{0}^{2 \pi} \frac{x \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x=
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Correct option: C
π2\pi^2
Option C is correct because by using the property ∫02af(x)dx=∫0af(x)dx+∫0af(2a−x)dx\int_0^{2a} f(x) dx = \int_0^{a} f(x) dx + \int_0^{a} f(2a-x) dx and simplifying, we can find the value of the integral to be π2\pi^2.

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