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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 2 · Q30
(a) Find: ∫ (cos 2 x)/(( sin x+ cos x)^2) d x
(a)
Find: $\displaystyle \int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$
OR
(b)
Evaluate: $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{5 \sin x+3 \cos x}{\sin x+\cos x} \mathrm{~d} x$
Marking-scheme solution
\[\begin{aligned}
\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x & =\int \frac{\cos ^{2} x-\sin ^{2} x}{(\sin x+\cos x)^{2}} d x \\
& =\int \frac{\cos x-\sin x}{\sin x+\cos x} d x \\
& =\log |\sin x+\cos x|+C
\end{aligned}
\]
\[\begin{aligned}
& I=\int_{0}^{\pi / 2} \frac{5 \sin x+3 \cos x}{\sin x+\cos x} d x \\
& I=\int_{0}^{\pi / 2} \frac{5 \sin \left(\dfrac{\pi}{2}-x\right)+3 \cos \left(\dfrac{\pi}{2}-x\right)}{\sin \left(\dfrac{\pi}{2}-x\right)+\cos \left(\dfrac{\pi}{2}-x\right)} d x=\int_{0}^{\pi / 2} \frac{5 \cos x+3 \sin x}{\cos x+\sin x} d x \\
& \text { Adding (i) and (ii), we get } \\
& \left.2 I=\int_{0}^{\pi / 2} 8 d x \Rightarrow I=4 x\right]_{0}^{\frac{\pi}{2}}=2 \pi
\end{aligned}
\]
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.