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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 26 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 26 Maths Textbook Solution.

Answers (1)

Answer:\pi -2

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

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

Solution:\int_{0}^{\frac{\pi }{2}}x^{2}\sin xdx

Integrating by parts =>Let  x^{2} be the first part and sin x be the 2nd part

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

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

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

(Again using integration by parts method)

\begin{array}{ll} =2\left[\frac{\pi}{2} \sin \frac{\pi}{2}-0 \times \sin 0\right]-2[-\cos x]_{0}^{\frac{\pi}{2}} & {\left[\int \sin x d x=-\cos x\right]} \\ =2\left[\frac{\pi}{2} \times 1-0\right]+2[\cos x]_{0}^{\frac{\pi}{2}} & {\left[\sin \frac{\pi}{2}=1, \sin 0=0\right]} \end{array}

\begin{aligned} &=\pi+2\left[\cos \frac{\pi}{2}-\cos 0\right] \\ &=\pi+2[0-1] \\ &=\pi-2 \end{aligned}                                                                            \left[\begin{array}{c} \cos \frac{\pi}{2}=0 \\ \cos 0=1 \end{array}\right]

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