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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 2 · Set 1 · Q27
(a) Evaluate: ∫_((π)/12)^((5 π)/12) (d x)/(1+√(cot x))
(a)
Evaluate: $\displaystyle \int_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \frac{\mathrm{~d} x}{1+\sqrt{\cot x}}$
OR
(b)
Evaluate : $\displaystyle \int_{\frac{-\pi}{6}}^{\frac{\pi}{2}}(\sin |x|+\cos |x|) \mathrm{d} x$
Marking-scheme solution
$\displaystyle I=\int_{\pi/12}^{5\pi/12} \dfrac{dx}{1+\sqrt{\cot x}}=\int_{\pi/12}^{5\pi/12} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, dx$ ... (i)
Applying $\displaystyle \int_{a}^{b} f(x)\, dx=\int_{a}^{b} f(a+b-x)\, dx$, we get
$\displaystyle I=\int_{\pi/12}^{5\pi/12} \dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\, dx$ ... (ii)
Adding (i) and (ii)
$\displaystyle 2I=\int_{\pi/12}^{5\pi/12} 1 \cdot dx$
$\displaystyle =[x]_{\pi/12}^{5\pi/12}=\dfrac{\pi}{3}$
$\displaystyle \Rightarrow I=\dfrac{\pi}{6}$
$\displaystyle I=\int_{-\pi/6}^{\pi/2}\left(\sin |x|+\cos |x|\right) dx$
$\displaystyle =\int_{-\pi/6}^{0}(-\sin x+\cos x)\, dx+\int_{0}^{\pi/2}(\sin x+\cos x)\, dx$
$\displaystyle =[\cos x+\sin x]_{-\pi/6}^{0}+[-\cos x+\sin x]_{0}^{\pi/2}$
$\displaystyle =\left(1-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)+(1+1)$
$\displaystyle =\dfrac{1}{2}(7-\sqrt{3})$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.