CBSE 2026 · Region 3 · Set 2 · Q31 · 3 marks
Evaluate : \[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} \mathrm{x}}{2^{\mathrm{x}}+1} d \mathrm{x} \]
Marking-scheme solution
Let, $\displaystyle I=\int_{-\pi / 2}^{\pi / 2} \dfrac{\cos^{2} x}{2^{x}+1} dx \quad \ldots(1)$
Using property $\displaystyle \int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$, we get
$\displaystyle I=\int_{-\pi / 2}^{\pi / 2} \dfrac{2^{x} \cos^{2} x}{2^{x}+1} dx \quad \ldots(2)$
On adding equations ($\displaystyle 1$) and ($\displaystyle 2$), we get
$\displaystyle 2 I=\int_{-\pi / 2}^{\pi / 2} \cos^{2} x\, dx=2 \int_{0}^{\pi / 2} \cos^{2} x\, dx \Rightarrow I=\int_{0}^{\pi / 2} \cos^{2} x\, dx \quad \ldots(3)$
Getting $\displaystyle I=\int_{0}^{\pi / 2} \sin^{2} x\, dx \quad \ldots(4)$
On adding equations ($\displaystyle 3$) and ($\displaystyle 4$), we get
$\displaystyle 2 I=\int_{0}^{\pi / 2} 1\, dx \Rightarrow I=\dfrac{\pi}{4}$
IntegralsSome Properties of Definite IntegralsApplyshort_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.