✓ Board-verified
Mathematics · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 1 · Q28
(a) Find: ∫ (x+ sin x)/(1+ cos x) d x
(a)
Find: $\displaystyle \int \frac{x+\sin x}{1+\cos x} d x$
OR
(b)
Evaluate: $\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{d x}{\cos ^{3} x \sqrt{2 \sin 2 x}}$
Marking-scheme solution
(a)
$\displaystyle \int \frac{x+\sin x}{1+\cos x}\,dx$. Using $\displaystyle 1+\cos x = 2\cos^2\frac{x}{2}$ and $\displaystyle \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$: the integrand $\displaystyle = \frac{x}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2}$. Integrating the first term by parts against $\displaystyle \int \tan\frac{x}{2}$ cancels the $\displaystyle \tan$ terms, giving $\displaystyle \boxed{x\tan\frac{x}{2} + C}$.
More from Integrals
- Evaluate: ∫_0^π x/(1+ sin x) d x2022 · asked 2×
- Evaluate ∫_-1^1|x^4-x| d x. OR Find ∫ (sin^-1 x)/((1-x^2)^3 / 2) d x.2023
- Find: ∫ (x^2)/((x^2+4)(x^2+9)) d x2023
- Evaluate: ∫_0^π (sin^2026 x)/(sin^2026 x+ cos^2026 x) d x2026
- Evaluate: ∫_1^3 (√x)/(√x+√(4-x)) dx2022
- Find: ∫ ((3 cos x-2) sin x)/(5- sin^2 x-4 cos x) d x OR Evaluate: ∫_-2^2 (x^3+|x|+1)/(x^2+4|x|+4) d x2024
- ∫ (x+5)/((x+6)^2) e^x dx is equal to:2025
- Evaluate ∫ 1/((e^x+e^-x)(e^x-e^-x)) d x log √22023
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.