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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 28 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 28 Maths Textbook Solution.

Answers (1)

Answer: \frac{\pi ^{2}}{4}-2

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

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

Solution:

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

Integrating by parts then x^{2} and cos x be the two parts

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

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

(Again using integrating by parts method)

\left[\begin{array}{l} \int \sin x d x=-\cos x \\ \sin \frac{\pi}{2}=1, \sin 0=0 \end{array}\right]

\begin{aligned} &=\frac{\pi^{2}}{4}+2[x \cos x]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} \cos x d x\\ &=\frac{\pi^{2}}{4}+2\left[\frac{\pi}{2} \cos \frac{\pi}{2}-0 \cos 0\right]-2[\sin x]_{0}^{\frac{\pi}{2}} \end{aligned}                                                            \left [ \int \cos xdx=\sin x \right ]

=\frac{\pi^{2}}{4}+2\left[\frac{\pi}{2} 0-0 \times 1\right]-2\left[\sin \frac{\pi}{2}-\sin 0\right] \quad\left[\begin{array}{l} \cos \frac{\pi}{2}=0 \\ \cos 0=1 \end{array}\right]

\begin{aligned} &=\frac{\pi^{2}}{4}+2 \times 0-2(1-0) \\ &=\frac{\pi^{2}}{4}-2 \end{aligned}

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