#### Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 56

Answer:  $I=\frac{\pi}{2}-2 \log \sqrt{2}$

Hint: To this equation convert cot into tan

Given:  $\int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right) d x$

Solution:

\begin{aligned} &\int_{0}^{1} \tan ^{-1}\left(\frac{1}{1-x+x^{2}}\right) d x \\ & \end{aligned}

$\int_{0}^{1} \tan ^{-1}\left(\frac{1}{(1-x)(1-x)}\right) d x \\$

$=\int_{0}^{1} \tan ^{-1}\left(\frac{x+1-x}{(1-x)(1-x)}\right)$

\begin{aligned} &=\int_{0}^{1}\left(\tan ^{-1} x+\tan ^{-1}(1-x)\right) d x \\ & \end{aligned}

$=\int_{0}^{1} \tan ^{-1} x d x+\int_{0}^{1} \tan ^{-1}(1-x) d x$

$\tan^{-1}x=t$

Put $x=\tan t$

\begin{aligned} &1=\sec ^{2} \frac{d t}{d x} \\ & \end{aligned}

$d x=\sec ^{2} d t \\$

$I=\int \tan ^{-1} x d x \\$

$=\int t \sec ^{2} t d t$

\begin{aligned} &=t \int \sec ^{2} t d t-\int \tan t d t \\ & \end{aligned}

$=t \tan t-\int \tan t d t \\$

$=t \tan t-\log |\sec t|$

\begin{aligned} &I_{2}=\int \tan ^{-1}(1-x) d x\\ & \end{aligned}

$\text { Put }$  $1-x=m\\$

$-1=\frac{d m}{d x}\\$

$d x=-d m\\$

$\text { Put }$

\begin{aligned} &1=\sec ^{2} \frac{v d v}{d m} \\ & \end{aligned}

$m=\tan v \\$

$d m=\sec ^{2} v d v \\$

$I_{2}=-\int \tan ^{-1} m d n \\$

$=-\int v \cdot \sec ^{2} v d v$

\begin{aligned} &=(v \tan v-\log |\sec v|) \\ & \end{aligned}

$\int \cot ^{-1}\left(1-x+x^{2}\right) d x=I_{1}+I_{2} \\$

$=t \tan t-\log \sec t|-v \tan v-\log | \sec v \mid \\$

$=t \tan t-\log \log |\sec t|-v \tan v+\log |\sec v| \\$

$t=\tan ^{-1} x \\$

$\tan t=x \\$

$\tan v=m=1-x$

\begin{aligned} &=\left[x \tan ^{-1}-\log \sqrt{1+x^{2}}-(1-x) \tan ^{-1}(1-x)+\log \left(\sqrt{1-x^{2}+1}\right]_{0}^{1}\right.\\ & \end{aligned}

$=\left(\frac{\pi}{4}-\log \sqrt{2}\right)-\left(\frac{-\pi}{4}+\log \sqrt{2}\right)$

\begin{aligned} &=\frac{\pi}{4}-\log \sqrt{2}+\frac{\pi}{4}-\log \sqrt{2} \\ & \end{aligned}

$=\frac{\pi}{2}-2 \log \sqrt{2}$