CBSE 2023 · Region 3 · Set 3 · Q28 · 3 marks
Evaluate : \[\int_{0}^{\pi / 4} \log (1+\tan \mathrm{x}) d \mathrm{x} \]
Marking-scheme solution
Let $\displaystyle \mathrm{I}=\int_{0}^{\pi / 4} \log (1+\tan \mathrm{x}) \mathrm{dx}$\begin{aligned}
& =\int_{0}^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-\mathrm{x}\right)\right) d \mathrm{x} \quad\left[\text { Using } \int_{0}^{a} f(\mathrm{x}) d \mathrm{x}=\int_{0}^{a} f(a-\mathrm{x}) d \mathrm{x}\right]
& =\int_{0}^{\pi / 4} \log \left(1+\frac{1-\tan \mathrm{x}}{1+\tan \mathrm{x}}\right) \mathrm{dx}
& =\int_{0}^{\pi / 4} \log \frac{2}{1+\tan \mathrm{x}} \mathrm{dx}
& 2 \mathrm{I}=\int_{0}^{\pi / 4} \log 2 d \mathrm{x}=\frac{\pi}{4} \log 2
& \mathrm{I}=\frac{\pi}{8} \log 2
\end{aligned}
$$
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.