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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 29 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 29 Maths Textbook Solution.

Answers (1)

Answer: \sqrt{2}+\frac{\pi }{2\sqrt{2}}-\frac{\pi ^{2}}{16\sqrt{2}}-2

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

Given: \int_{0}^{\frac{\pi }{4}}x^{2}\sin xdx

Solution:

\int_{0}^{\frac{\pi }{4}}x^{2}\sin xdx

Integrating by parts then, x^{2} and sin x be the 2 parts

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

=-\left[x^{2} \cos x\right]_{0}^{\frac{\pi}{4}}+2 \int_{0}^{\frac{\pi}{4}} x \cos x d x

(Again using integration by parts method x and cos x be the 2 terms)

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

=-\left[\frac{\pi^{2}}{16} \frac{1}{\sqrt{2}}-0\right]+2[x \sin x]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} 1 \cdot \sin x d \quad\left[\begin{array}{l} \int \cos x d x=\sin x \\ \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1 \end{array}\right]

\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4} \sin \frac{\pi}{4}-0 \times \sin 0\right]-2 \int_{0}^{\frac{\pi}{4}} \sin x d x \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4} \frac{1}{\sqrt{2}}-0\right]-2[-\cos x]_{0}^{\frac{\pi}{4}} \end{aligned}\left[\begin{array}{l} \int \sin x d x=\cos x \\ \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 0=0 \end{array}\right]

\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4 \sqrt{2}}\right]+2[\cos x]_{0}^{\frac{\pi}{4}} \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+2\left[\cos \frac{\pi}{4}-\cos 0\right] \quad\left[\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1\right] \end{aligned}

\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+2\left[\frac{1}{\sqrt{2}}-1\right] \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}-2 \\ &=\sqrt{2}+\frac{\pi}{2 \sqrt{2}}-\frac{\pi^{2}}{16 \sqrt{2}}-2 \end{aligned}

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