CBSE 2023 · Region 2 · Set 1 · Q26 · 3 marks
Evaluate $\displaystyle \int_{0}^{\pi / 4} \log (1+\tan \mathrm{x}) \mathrm{d} \mathrm{x}$.Find $\displaystyle \int \frac{\mathrm{d} \mathrm{x}}{\sqrt{\sin ^{3} \mathrm{x} \cos (\mathrm{x}-\alpha)}}$.
Evaluate $\displaystyle \int_{0}^{\pi / 4} \log (1+\tan \mathrm{x}) \mathrm{d} \mathrm{x}$.
Find $\displaystyle \int \frac{\mathrm{d} \mathrm{x}}{\sqrt{\sin ^{3} \mathrm{x} \cos (\mathrm{x}-\alpha)}}$.
Marking-scheme solution
(a)
Let $\displaystyle \mathrm{I}=\int_{0}^{\pi / 4} \log (1+\tan \mathrm{x}) \mathrm{dx}$
$$\Rightarrow \mathrm{I}=\int_{$\displaystyle 0$}^{\pi / $\displaystyle 4$} \log \left($\displaystyle 1$+\tan \left(\frac{\pi}{4}-\mathrm{x}\right)\right) \mathrm{d} \mathrm{x}=\int_{$\displaystyle 0$}^{\pi / $\displaystyle 4$} \log \left(\frac{2}{$\displaystyle 1$+\tan \mathrm{x}}\right) \mathrm{d} \mathrm{x}
$$Add (i) and (ii), $\displaystyle \left.2 \mathrm{I}=\int_{0}^{\pi / 4} \log 2 \mathrm{dx}=\log 2 \cdot \mathrm{x}\right]_{0}^{\pi / 4}=\frac{\pi}{4} \log 2$
$$\Rightarrow \mathrm{I}=\frac{\pi}{8} \log $\displaystyle 2$
$$Or
(b) Let $\displaystyle \mathrm{I}=\int \frac{1}{\sqrt{\sin ^{3} \mathrm{x} \cos (\mathrm{x}-\alpha)}} \mathrm{dx}=\int \frac{\operatorname{cosec}^{2} \mathrm{x}}{\sqrt{\sin \alpha+\cos \alpha \cot \mathrm{x}}} \mathrm{dx}$
$$\begin{gathered}
\text { Put } \sin \alpha+\cos \alpha \cot \mathrm{x}=t \Rightarrow \operatorname{cosec}^{$\displaystyle 2$} \mathrm{x} \mathrm{d} \mathrm{x}=-\frac{1}{\cos \alpha} \mathrm{d} t
\therefore \mathrm{I}=-\int \frac{1}{\cos \alpha \sqrt{t}} \mathrm{d} t=-\frac{$\displaystyle 2$ \sqrt{t}}{\cos \alpha}+c
\Rightarrow \mathrm{I}=-\frac{2}{\cos \alpha} \sqrt{\sin \alpha+\cos \alpha \cot \mathrm{x}}+c
\end{gathered}
IntegralsSome Properties of Definite IntegralsApplyshort_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.