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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 3 · Set 1 · Q28
(a) Find: ∫ (cos x)/((2+ sin x )(4+ sin x )) d x
(a)
Find : \[\int \frac{\cos \mathrm{x}}{(2+\sin \mathrm{x})(4+\sin \mathrm{x})} d \mathrm{x} \]
OR
(b)
Find : \[\int \frac{\mathrm{x}+3}{\mathrm{x}^{2}+4 \mathrm{x}+5} d \mathrm{x} \]
Marking-scheme solution
(a)
Put $\displaystyle \sin x=t \Rightarrow \cos x\, dx=dt$
$\displaystyle \int \dfrac{\cos x}{(2+\sin x)(4+\sin x)} dx=\int \dfrac{1}{(2+t)(4+t)} dt$
$\displaystyle =\dfrac{1}{2} \int\left(\dfrac{1}{2+t}-\dfrac{1}{4+t}\right) dt=\dfrac{1}{2} \log\left|\dfrac{2+t}{4+t}\right|+c$
$\displaystyle =\dfrac{1}{2} \log\left|\dfrac{2+\sin x}{4+\sin x}\right|+c$
OR
(b)
$\displaystyle \int \dfrac{x+3}{x^{2}+4 x+5} dx=\dfrac{1}{2} \int \dfrac{(2 x+4)+2}{x^{2}+4 x+5} dx$
$\displaystyle =\dfrac{1}{2} \int \dfrac{2 x+4}{x^{2}+4 x+5} dx+\int \dfrac{1}{x^{2}+4 x+5} dx$
$\displaystyle =\dfrac{1}{2} \log|x^{2}+4 x+5|+\int \dfrac{1}{(x+2)^{2}+1} dx$
$\displaystyle =\dfrac{1}{2} \log|x^{2}+4 x+5|+\tan^{-1}(x+2)+c$
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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.