(1+)/(1-) d = Option 1: ()/2 + ()/2 + c, where c is an arbitrary constant Option 2: ()/2 + + c, where c is an arbitrary constant Option 3: (()/2) + + c, where c is an arbitrary constant Option 4: -2(()/2) - + c, where c is an arbitrary constant

Answer: -2(()/2) - + c, where c is an arbitrary constant. Solution: Using half-angle formulas, (1+)/(1-) = ^2()/2 = ^2()/2 - 1. Integrating gives -2()/2 - + c.

Calculus

Integrals

Medium

core

$\displaystyle\int \dfrac{1+\cos\theta}{1-\cos\theta}\,d\theta =$

$\tan\dfrac{\theta}{2} + \sec\dfrac{\theta}{2} + c$ , where $c$ is an arbitrary constant

$\tan\dfrac{\theta}{2} + \theta + c$ , where $c$ is an arbitrary constant

$\cot\left(\dfrac{\theta}{2}\right) + \theta + c$ , where $c$ is an arbitrary constant

$-2\cot\left(\dfrac{\theta}{2}\right) - \theta + c$ , where $c$ is an arbitrary constant

✅ Correct Option: 4

Using half-angle formulas, $\frac{1+\cos\theta}{1-\cos\theta} = \cot^2\frac{\theta}{2} = \csc^2\frac{\theta}{2} - 1$ . Integrating gives $-2\cot\frac{\theta}{2} - \theta + c$ .

2025: 30 May Shift 2

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)dx$ , is equal to:

2022: 23 Aug Shift 1

$\int \frac{x^2 - 4}{(x+2)(x-1)(x-3)} dx =$

2025: 19 May Shift 1

The value of the definite integral $I = \int_0^2 x\sqrt{2-x} dx$ is:

2025: 22 May Shift 2

$\int \frac{\sin 2x \, dx}{\sqrt{9 - \cos^4 x}}$ equals

2022: 16 July Shift 2

$\int \frac{x}{(x-1)(x-2)} dx =$

(where c is an arbitrary constant)

2026: 22 May Shift 1

Match List-I with List-II

(Consider $\log_e x = \log x$ )

List-I	List-II
Integral	Solution: where C is an arbitrary constant
(A) $\displaystyle\int \dfrac{f'(x)}{f(x)\log f(x)}\,dx$	(I) $x\log(\log x) + c$
(B) $\displaystyle\int \left(\log(\log x) + \dfrac{1}{\log x}\right) dx$	(II) $\dfrac{x}{\log x} + C$
(C) $\displaystyle\int \dfrac{\log x}{(1+\log x)^2}\,dx$	(III) $\log\left\vert \log(f(x)) \right\vert + c$
(D) $\displaystyle\int \left(\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right) dx$	(IV) $\dfrac{x}{1+\log x} + c$

Choose the correct answer from the options given below:

∫1+cos⁡θ1−cos⁡θ dθ=\displaystyle\int \dfrac{1+\cos\theta}{1-\cos\theta}\,d\theta =∫1−cosθ1+cosθ​dθ=

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CUET Mathematics Past Year Questions (Topic-Wise)

$\displaystyle\int \dfrac{1+\cos\theta}{1-\cos\theta}\,d\theta =$