Evaluate: \int_{-\pi}^{\pi} (1-x^2)\sin x \cos^2 x dx.

 

 

 

 
 
 
 
 

Answers (1)

I = \int_{-\pi}^{\pi} (1-x^2)\sin x \cos^2 x dx

Let f(x)= (1-x^2)\sin x \cos^2 x

       f(-x)= - (1-x^2)\sin x \cos^2 x\quad \begin{bmatrix} \because \cos(-\theta )= \cos \theta\\ \sin (-\theta) = - \sin\theta \end{bmatrix}

            \therefore f(-x) = - f(x)

So, f(x) is odd

            \therefore \int_{-a}^{a}f(x)dx=0

            \therefore I=0

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