JEE MainMathematicsEasyDefinite Integration

$\int_{- \pi/2}^{\pi/2}\mspace{2mu}\sin^{10}x\left( 6x^{9} - 25x^{7} + 4x^{3} - 2x \right)$
dx equals.

Show answer and solution

Let f(x) = sin^10 x (6x^9 - 25x^7 + 4x^3 - 2x)

\Rightarrow

f(-x) = sin^10 x (- 6x^9 + 25x^7 - 4x^3 + 2x) = -
f(x)

\Rightarrow

f(x) is an odd function

\therefore

\int_{- \pi/2}^{\pi/2}\mspace{2mu}\sin^{10}x\left( 6x^{9} - 25x^{7} + 4x^{3} - 2x \right)dx = 0

2\mathcal{l}np = \frac{2}{\pi}\int_{0}^{\pi/2}\mspace{2mu}\mathcal{(l}n(tant) + \mathcal{l}n(cott)dt

= \frac{2}{\pi}\int_{0}^{\pi/2}\mspace{2mu}\mathcal{l}n(tantcott)dt = 0

p = 1

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

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