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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.

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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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