CBSE 2026 · Region 1 · Set 2 · Q30 · 3 marks
Evaluate : $\displaystyle \int_{0}^{1} x \sin ^{-1} x \mathrm{~d} x$
Marking-scheme solution
$\displaystyle \int_0^1 x\sin^{-1}x\,dx = \left[\sin^{-1}x \times \dfrac{x^2}{2}\right]_0^1 - \int_0^1 \dfrac{1}{\sqrt{1-x^2}}\times\dfrac{x^2}{2}\,dx$
$\displaystyle = \dfrac{\pi}{4} + \int_0^1 \dfrac{1-x^2-1}{\sqrt{1-x^2}}\times\dfrac{1}{2}\,dx$
$\displaystyle = \dfrac{\pi}{4} + \dfrac{1}{2}\left[\int_0^1 \sqrt{1-x^2}\,dx - \int_0^1 \dfrac{1}{\sqrt{1-x^2}}\,dx\right]$
$\displaystyle = \dfrac{\pi}{4} + \dfrac{1}{2}\left[\dfrac{x}{2}\sqrt{1-x^2} + \dfrac{1}{2}\sin^{-1}x - \sin^{-1}x\right]_0^1$
$\displaystyle = \dfrac{\pi}{4} + \dfrac{1}{2}\left[\dfrac{x}{2}\sqrt{1-x^2} - \dfrac{1}{2}\sin^{-1}x\right]_0^1$
$\displaystyle = \dfrac{\pi}{4} - \dfrac{\pi}{8} = \dfrac{\pi}{8}$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.