CBSE 2025 · Region 2 · Set 2 · Q22 · 2 marks
Evaluate: $\displaystyle \int_{0}^{\pi} \frac{\sin 2 \mathrm{p} x}{\sin x} \mathrm{~d} x, \mathrm{p} \in \mathrm{N}$.
Marking-scheme solution
\[\begin{aligned}
& \mathrm{I}=\int_{0}^{\pi} \frac{\sin 2 \mathrm{p} x}{\sin x} d x \\
& =\int_{0}^{\pi} \frac{\sin 2 \mathrm{p}(\pi-x)}{\sin (\pi-x)} d x \\
& \mathrm{I}=\int_{0}^{\pi} \frac{-\sin 2 \mathrm{p} x}{\sin x} d x
\end{aligned}
\]
Adding, we get
$\displaystyle 2 \mathrm{I}=0$
$\displaystyle \therefore \mathrm{I}=0$
IntegralsSome Properties of Definite IntegralsApplyvery_short_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.