Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
Evaluate: $\int_0^1 x \cdot \tan ^{-1} x d x$
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Answer
$\text { Let } I =\int_0^1 x \tan ^{-1} x d x$
$=\left[\tan ^{-1} x \int x d x\right]_0^1-\int_0^1\left[\frac{ d }{ d x}\left(\tan ^{-1} x\right) \int x d x\right] d x$
$=\left[\tan ^{-1} x \cdot \frac{x^2}{2}\right]_0^1-\int_0^1 \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x$
$=\left[\frac{x^2}{2} \tan ^{-1} x\right]_0^1-\frac{1}{2} \int_0^1 \frac{x^2}{1+x^2} d x$
$=\left[\frac{1}{2} \tan ^{-1}-0\right]_{-\frac{1}{2}} \frac{x^2+1-1}{1+x^2} d x$
$=\frac{1}{2} \cdot \frac{\pi}{4}-\frac{1}{2} \int_0^1\left(1-\frac{1}{1+x^2}\right) d x$
$=\frac{\pi}{8}-\frac{1}{2}\left[x-\tan ^{-1} x\right]_0^1$
$=\frac{\pi}{8}-\frac{1}{2}\left[\left(1-\tan ^{-1} 1\right)-\left(0-\tan ^{-1} 0\right)\right]$ $=\frac{\pi}{8}-\frac{1}{2}\left(1-\frac{\pi}{4}-0\right)$
$=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}$
$\therefore I =\frac{\pi}{4}-\frac{1}{2}$
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