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  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 32 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 32 maths.

Answers (1)

Answer: \frac{\pi}{4}-\frac{1}{2}

Hint: We use indefinite formula then put limits to solve this integral.

Given: \int_{0}^1{} x\tan ^{-1}xdx

Solution: \int_{0}^1{} x\tan ^{-1}xdx

Applying integration by parts, then,

\int_{0}^1{} x\tan ^{-1}xdx

\begin{aligned} &=\left[\tan ^{-1} x \int x d x\right]_{0}^{1}-\int_{0}^{1}\left\{\frac{d}{d x}\left(\tan ^{-1} x\right) \int x d x\right\} d x \\ &=\left[\tan ^{-1} x \cdot \frac{x^{1+1}}{1+1}\right]_{0}^{1}-\int_{0}^{1} \frac{1}{1+x^{2}} \cdot \frac{x^{2}}{2} \mathrm{dx} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} \frac{x^{2}}{1+x^{2}} \mathrm{dx} \\ &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1}\left(\frac{x^{2}+1-1}{1+x^{2}}\right) \mathrm{dx} \end{aligned}

\begin{aligned} &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1}\left(\frac{1+x^{2}}{1+x^{2}}-\frac{1}{1+x^{2}}\right) \mathrm{d} x \\ &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} 1 d x+\frac{1}{2} \int \frac{1}{1+x^{2}} d x \\ &=\left[\tan ^{-1} 1 \cdot \frac{1^{2}}{2}-\frac{0^{2}}{2} \cdot \tan ^{-1} 0\right]-\frac{1}{2}[x]_{0}^{1}+\frac{1}{2}\left[\tan ^{-1} x\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\left[\frac{1}{2} \cdot \tan ^{-1} 1-0\right]-\frac{1}{2}[1-0]+\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\ &=\frac{1}{2} \cdot \frac{\pi}{4}-\frac{1}{2} \cdot 1+\frac{1}{2}\left[\frac{\pi}{4}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan ^{-1} 1=\frac{\pi}{4}, \tan ^{-1} 0=0\right] \\ &=\frac{1}{2} \cdot \frac{\pi}{4}-\frac{1}{2}+\frac{1}{2}, \frac{\pi}{4} \\ &=\frac{\pi}{4}\left(\frac{1}{2}+\frac{1}{2}\right)-\frac{1}{2} \\ &=\frac{\pi}{4}-\frac{1}{2} \end{aligned}

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