CBSE 2024 · Region 5 · Set 1 · Q24 · 2 marks
Find : $\displaystyle \int \cos ^{3} x \mathrm{e}^{\log \sin x} \mathrm{~d} x$Find : $\displaystyle \int \frac{1}{5+4 x-x^{2}} \mathrm{~d} x$
Find : $\displaystyle \int \cos ^{3} x \mathrm{e}^{\log \sin x} \mathrm{~d} x$
Find : $\displaystyle \int \frac{1}{5+4 x-x^{2}} \mathrm{~d} x$
Marking-scheme solution
(a)
$\displaystyle \int \cos ^{3} x \mathrm{e}^{\log \sin x} d x=\int \cos ^{3} x \cdot \sin x d x$, Assuming $\displaystyle \cos x=t$ and $\displaystyle \sin x d x=-d t$
$$\begin{aligned}
& =-\int t^{$\displaystyle 3$} d t
& =-\frac{t^{$\displaystyle 4$}}{$\displaystyle 4$}+C=-\frac{\cos ^{$\displaystyle 4$} x}{$\displaystyle 4$}+C
\end{aligned}
$$Or
(b) $\displaystyle \int \frac{1}{5+4 x-x^{2}} d x=\int \frac{1}{3^{2}-(x-2)^{2}} d x$
$$=\frac{1}{6} \log \left|\frac{1+x}{5-x}\right|+C
IntegralsIntegrals of Some Particular FunctionsApplyvery_short_answermedium
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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.