CBSE 2022 · Region 2 · Set 1 · Q2 · 2 marks
Find : \[\int \frac{\sin 3 \mathrm{x}}{\sin \mathrm{x}} d \mathrm{x} \]Evaluate : \[\int_{0}^{\frac{1}{2} \log 3} \frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{2 \mathrm{x}}+1} \mathrm{dx} \]
Find : \[\int \frac{\sin 3 \mathrm{x}}{\sin \mathrm{x}} d \mathrm{x} \]
Evaluate : \[\int_{0}^{\frac{1}{2} \log 3} \frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{2 \mathrm{x}}+1} \mathrm{dx} \]
Marking-scheme solution
(a)
\[\int \frac{\sin 3x}{\sin x}\,dx = \int \frac{3\sin x - 4\sin^3 x}{\sin x}\,dx\]
\[= \int \left[3 - 4\frac{(1-\cos 2x)}{2}\right] dx\]
\[= \int (1 + 2\cos 2x)\,dx\]
\[= x + \sin 2x + C\]
Or(b) Let $\displaystyle e^x = t$, $\displaystyle e^x dx = dt$
\[\therefore \int_{0}^{\frac{1}{2}\log 3} \frac{e^x}{e^{2x}+1}\,dx = \int_{1}^{\sqrt{3}} \frac{dt}{t^2+1}\]
\[= \tan^{-1} t \Big|_{1}^{\sqrt{3}}\]
\[= \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}\]
IntegralsMethods of IntegrationApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.