#### Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 23.

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

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

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

Solution:$I=\int_{0}^{1}\tan ^{-1}x dx$

Applying integration by parts method we get

$I=\int_{0}^{1}\tan ^{-1}x dx$

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

Put  $1+x^2=t$

$2x \; \; dx=dt$

$x \; dx=\frac{dt}{2}$

Then

\begin{aligned} &I=\left[\tan ^{-1} x \cdot x\right]_{0}^{1}-\int_{0}^{-1} \frac{1 d t}{2} \\ &=\left[1 \cdot \tan ^{-1} 1-0 \cdot \tan ^{-1} 0\right]-\frac{1}{2}[\log |t|]_{0}^{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log |x|\right] \\ &=\left[\frac{\pi}{4}-0\right]-\frac{1}{2}\left[\log \left|1+x^{2}\right|\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\frac{\pi}{4}-\frac{1}{2}\left[\log \left|1+1^{2}\right|-\log \left|1+0^{2}\right|\right] \\ &=\frac{\pi}{4}-\frac{1}{2}[\log (1+1)-\log 1] \\ &=\frac{\pi}{4}-\frac{1}{2}[\log 2-\log 1] \end{aligned}

$=\frac{\pi}{4}-\frac{1}{2}[\log 2-0]$                                                        $[\log 1 =0]$

$=\frac{\pi}{4}-\frac{1}{2}[\log 2]$