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Question: The value of \(\int_{- \pi}^{\pi}{(1 - x^{2})\sin x\cos^{2}xdx}\) is...

The value of ∫−ππ(1−x2)sin⁡xcos⁡2xdx\int_{- \pi}^{\pi}{(1 - x^{2})\sin x\cos^{2}xdx} is

A

0

B

1

C

2

D

3

Answer

0

Explanation

Solution

Let, f1(x)=(1−x2)f_{1}(x) = (1 - x^{2}), f2(x)=sin⁡xf_{2}(x) = \sin x and f3(x)=cos⁡2xf_{3}(x) = \cos^{2}x

Now, f1(x)=f1(−x)f_{1}(x) = f_{1}( - x), f2(x)=−f2(−x)f_{2}(x) = - f_{2}( - x) and f3(x)=f(−x)f_{3}(x) = f( - x)

∴ I=∫−ππf(x)dx=∫−ππ[f1(x).f2(x).f3(x)]dxI = \int_{- \pi}^{\pi}{f(x)dx = \int_{- \pi}^{\pi}{\lbrack f_{1}(x).f_{2}(x).f_{3}(x)\rbrack dx}}

= −∫−ππ[f1(−x).f2(−x).f3(−x)]dx- \int_{- \pi}^{\pi}{\lbrack f_{1}( - x).f_{2}( - x).f_{3}( - x)\rbrack dx}

∴ I=0I = 0

(6)∫02af(x)dx=∫0af(x)dx+∫0a f(2a−x)dx\int_{0}^{2a}{f(x)dx} = \int_{0}^{a}{f(x)dx + \int_{0}^{a}\ f(2a - x)dx}In particular, ∫02af(x)dx={0,iff(2a−x)=−f(x)2∫0af(x)dx,iff(2a−x)=f(x) \int_{0}^{2a}{}f(x)dx = \left\{ \begin{matrix} 0,\text{if}f(2a - x) = - f(x) \\ 2\int_{0}^{a}{f(x)dx},\text{if}f(2a - x) = f(x) \end{matrix} \right.\ It is generally used to make half the upper limit.