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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 30 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 30 Maths Textbook Solution.

Answers (1)

Answer: \frac{-\pi }{4}

Hint: Use indefinite formula then put the limit to solve this integral

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

Solution:

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

Integrating by parts then

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

=\frac{1}{2}\left[x^{2} \sin 2 x\right]-\frac{2}{2} \int_{0}^{\frac{\pi}{2}} x \sin 2 x d x

Again using integrating on by parts method

=\frac{1}{2}\left[\left(\frac{\pi}{2}\right)^{2} \sin 2 \times \frac{\pi}{2}-0^{2} \sin 0\right]-\left[x \int \sin 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{d(x)}{d x} \int \sin 2 x d x\right) d x

\begin{aligned} &=\frac{1}{2}\left[\frac{\pi^{2}}{4} \sin \pi-\sin 0 \times 0\right]-\left[x\left(\frac{-\cos 2 x}{2}\right)\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\left(\frac{-\cos 2 x}{2}\right) d x \\ &=\frac{1}{2} \times 0+\frac{1}{2}[x \cos 2 x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \frac{\cos 2 x}{2} d x & {\left[\begin{array}{l} \int \sin a x d x=\frac{-\cos a x}{a} \\ \sin \pi=0, \sin 0=0 \end{array}\right]} \end{aligned}

=\frac{1}{2}\left[\frac{\pi}{2} \cos 2 \times \frac{\pi}{2}-0 \times \cos 2 \times 0\right]-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \quad\left[\int \cos a x d x=\frac{\sin a x}{a}\right]

\begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{2} \cos \pi-0\right]-\frac{1}{2 \times 2}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}\left[\frac{\pi}{2}\right](-1)-\frac{1}{4}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \end{aligned}                                                            \left [ \cos \pi =-1 \right ]

\begin{aligned} &=\frac{1}{2}\left(\frac{-\pi}{2}\right)-\frac{1}{4}[\sin \pi-\sin 0] \\ &=-\frac{\pi}{4}-\frac{1}{4}[0-0] \end{aligned}                                                                            \left [ \sin \pi =\sin 0=0 \right ]

=\frac{-\pi }{4}

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