CBSE 2026 · Region 5 · Set 1 · Q27 · 3 marks
Find : \[\int \frac{\mathrm{x}-\sin \mathrm{x}}{1-\cos \mathrm{x}} d \mathrm{x} \]Evaluate : \[\int_{0}^{2} \frac{1}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}} d \mathrm{x} \]
Find : \[\int \frac{\mathrm{x}-\sin \mathrm{x}}{1-\cos \mathrm{x}} d \mathrm{x} \]
Evaluate : \[\int_{0}^{2} \frac{1}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}} d \mathrm{x} \]
Marking-scheme solution
Let $\displaystyle I=\int \dfrac{x-\sin x}{1-\cos x} d x$
$\displaystyle =\int \dfrac{x}{1-\cos x} d x-\int \dfrac{\sin x}{1-\cos x} d x$
$\displaystyle =\int \dfrac{x}{2 \sin^{2} \frac{x}{2}} d x-\int \dfrac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^{2} \frac{x}{2}} d x$
$\displaystyle =-\dfrac{x}{2} \cdot 2 \cot \dfrac{x}{2}-\dfrac{1}{2} \int-2 \cot \dfrac{x}{2} d x-\int \cot \dfrac{x}{2} d x$
$\displaystyle =-x \cot \dfrac{x}{2}+C$
$\displaystyle I=\int_{0}^{2} \dfrac{1}{\sqrt{x^{2}+2 x+3}} d x$
$\displaystyle =\int_{0}^{2} \dfrac{1}{\sqrt{(x+1)^{2}+2}} d x$
$\displaystyle =\left[\log \left(x+1+\sqrt{x^{2}+2 x+3}\right)\right]_{0}^{2}$
$\displaystyle =\log (3+\sqrt{11})-\log (1+\sqrt{3})$ or $\displaystyle \log\left(\dfrac{3+\sqrt{11}}{1+\sqrt{3}}\right)$
IntegralsMethods of IntegrationApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.