CBSE 2023 · Region 4 · Set 1 · Q27 · 3 marks
Evaluate : \[\int_{\pi / 4}^{\pi / 2} e^{2 \mathrm{x}}\left(\frac{1-\sin 2 \mathrm{x}}{1-\cos 2 \mathrm{x}}\right) d \mathrm{x} \]Evaluate : \[\int_{-2}^{2} \frac{\mathrm{x}^{2}}{1+5^{\mathrm{x}}} d \mathrm{x} \]
Evaluate : \[\int_{\pi / 4}^{\pi / 2} e^{2 \mathrm{x}}\left(\frac{1-\sin 2 \mathrm{x}}{1-\cos 2 \mathrm{x}}\right) d \mathrm{x} \]
Evaluate : \[\int_{-2}^{2} \frac{\mathrm{x}^{2}}{1+5^{\mathrm{x}}} d \mathrm{x} \]
Marking-scheme solution
$$\text { (a) } \mathrm{I}=\int_{\boldsymbol{\pi} / \mathbf{4}}^{\boldsymbol{\pi} / \mathbf{2}} \boldsymbol{e}^{\mathbf{2} \boldsymbol{\mathrm{x}}}\left(\frac{1-\sin 2 \mathrm{x}}{\mathbf{1}-\boldsymbol{\operatorname { c o s }} 2 \mathrm{x}}\right) \mathrm{dx}Put $\displaystyle 2 \mathrm{x}=\mathrm{t}$ so that $\displaystyle 2 \mathrm{dx}=\mathrm{dt}$
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.