CBSE 2026 · Region 3 · Set 3 · Q28 · 3 marks
Evaluate : \[\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \mathrm{x}}{1+\sin 2 \mathrm{x}} d \mathrm{x} \]
Marking-scheme solution
Let, $\displaystyle I=\int_{0}^{\pi / 2} \dfrac{\sin^{2} x}{1+\sin 2x} dx \quad \ldots(1)$
Getting $\displaystyle I=\int_{0}^{\pi / 2} \dfrac{\cos^{2} x}{1+\sin 2x} dx \quad \ldots(2)$
On adding equations ($\displaystyle 1$) and ($\displaystyle 2$), we get
$\displaystyle 2 I=\int_{0}^{\pi / 2} \dfrac{1}{1+\sin 2x} dx$
$\displaystyle \Rightarrow 2 I=\int_{0}^{\pi / 2} \dfrac{1-\sin 2x}{\cos^{2} 2x} dx=\int_{0}^{\pi / 2} \sec^{2} 2x\, dx-\int_{0}^{\pi / 2} \sec 2x \tan 2x\, dx$
$\displaystyle \Rightarrow 2 I=\left[\dfrac{\tan 2x}{2}-\dfrac{\sec 2x}{2}\right]_{0}^{\pi / 2} \Rightarrow I=\dfrac{1}{2}$
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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.