CBSE 2024 · Region 5 · Set 1 · Q28 · 3 marks
Evaluate : $\displaystyle \int_{0}^{\pi} \frac{\mathrm{e}^{\cos \mathrm{x}}}{\mathrm{e}^{\cos \mathrm{x}}+\mathrm{e}^{-\cos \mathrm{x}}} \mathrm{~d} \mathrm{x}$Find : $\displaystyle \int \frac{2 \mathrm{x}+1}{(\mathrm{x}+1)^{2}(\mathrm{x}-1)} \mathrm{d} \mathrm{x}$
Evaluate : $\displaystyle \int_{0}^{\pi} \frac{\mathrm{e}^{\cos \mathrm{x}}}{\mathrm{e}^{\cos \mathrm{x}}+\mathrm{e}^{-\cos \mathrm{x}}} \mathrm{~d} \mathrm{x}$
Find : $\displaystyle \int \frac{2 \mathrm{x}+1}{(\mathrm{x}+1)^{2}(\mathrm{x}-1)} \mathrm{d} \mathrm{x}$
Marking-scheme solution
(a)
Let $\displaystyle \mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{e}^{\cos \mathrm{x}}}{\mathrm{e}^{\cos \mathrm{x}}+\mathrm{e}^{-\cos \mathrm{x}}} \mathrm{dx}$
$$\Rightarrow \mathrm{I}=\int_{$\displaystyle 0$}^{\pi} \frac{\mathrm{e}^{\cos (\pi-\mathrm{x})}}{\mathrm{e}^{\cos (\pi-\mathrm{x})}+\mathrm{e}^{-\cos (\pi-\mathrm{x})}} \mathrm{d} \mathrm{x}=\int_{$\displaystyle 0$}^{\pi} \frac{\mathrm{e}^{-\cos \mathrm{x}}}{\mathrm{e}^{-\cos \mathrm{x}}+\mathrm{e}^{\cos \mathrm{x}}} \mathrm{d} \mathrm{x}
$$Adding (i) and (ii), we get
$$\left.2 \mathrm{I}=\int_{$\displaystyle 0$}^{\pi} \mathrm{d} \mathrm{x}=\mathrm{x}\right]_{$\displaystyle 0$}^{\pi}=\pi, \quad \therefore \mathrm{I}=\frac{\pi}{2}
IntegralsSome Properties of Definite IntegralsApplyshort_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.