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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 21

Answers (1)

Answer:  \frac{-\pi}{4}

Given:   \int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x

Hint: Apply integration by parts method

Solution:   \int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x

\begin{aligned} &{\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x \frac{\sin 2 x}{2} d x} \\ & \end{aligned}

{\left[\because \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x\right]}

\begin{aligned} &=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\left[\left(-x \frac{\cos 2 x}{2}\right)+\int_{0}^{\frac{\pi}{2}} \frac{\cos 2 x}{2} d x\right] \\ \end{aligned}

=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}+\left(x \frac{\cos 2 x}{2}\right)_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left(\frac{\sin 2 x}{2}\right)_{0}^{\frac{\pi}{2}} \\

=0-\frac{\pi}{2} \times \frac{1}{2}-0=-\frac{\pi}{4}

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