Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{-1}^1 x \tan ^{-1} x d x$ equals
✓
$\frac{\pi}{2}-1$
B
$\frac{\pi}{2}+1$
C
$\pi-1$
D
$0$
✓
Answer
Correct option: A.
$\frac{\pi}{2}-1$
(A) Since $x \tan ^{-1} x$ is an even function. $\therefore \quad \int_{-1}^1 x \tan ^{-1} x d x=2 \int_0^1 x \tan ^{-1} x d x$ $\begin{array}{l}=\left[2 \tan ^{-1} x \cdot \frac{x^2}{2}\right]_0^1-2 \int_0^1 \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x \\ =\left[x^2 \tan ^{-1} x\right]_0^1-\int_0^1 \frac{x^2+1-1}{1+x^2} d x \\ =\left[x^2 \tan ^{-1} x\right]_0^1-[x]_0^1+\left[\tan ^{-1} x\right]_0^1 \\ =\frac{\pi}{4}-1+\frac{\pi}{4}=\frac{\pi}{2}-1\end{array}$
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