CBSE 2024 · Region 1 · Set 2 · Q32 · 5 marks
Evaluate : \[\int_{0}^{\pi / 2} e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x \]Find : \[\int_{\pi / 6}^{\pi / 3} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x \]
Evaluate : \[\int_{0}^{\pi / 2} e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x \]
Find : \[\int_{\pi / 6}^{\pi / 3} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x \]
Marking-scheme solution
(b)
$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
Put $\displaystyle \sin x-\cos x=t$ so that $\displaystyle (\cos x+\sin x) d x=d t$
$$=\int_{-\left(\frac{\sqrt{3}-$\displaystyle 1$}{$\displaystyle 2$}\right)}^{\frac{\sqrt{3}-$\displaystyle 1$}{$\displaystyle 2$}} \frac{d t}{\sqrt{$\displaystyle 1$-t^{$\displaystyle 2$}}}
$$$=\left[\sin ^{-1} t\right]_{-\left(\frac{\sqrt{3}-1}{2}\right)}^{\frac{\sqrt{3}-1}{2}}$
$$=$\displaystyle 2$ \sin ^{-$\displaystyle 1$}\left(\frac{\sqrt{3}-$\displaystyle 1$}{$\displaystyle 2$}\right)
IntegralsEvaluation of Definite Integrals by SubstitutionApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.