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

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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


\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


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}

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