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

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

Answers (1)

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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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