If 1/( x + x)dx = k 2x + C, where C is an arbitrary constant, then the value of k is Option 1: 1/2 Option 2: -1/2 Option 3: 1/4 Option 4: -1/4

Answer: -1/4. Solution: x+ x = 1/( x x), so the integrand is x x = 1/2 2x. 1/2 2x dx = -1/4 2x+C, so k = -1/4.

Calculus

Integrals

Easy

core

If $\int \frac{1}{\tan x + \cot x}dx = k\cos 2x + C$ , where C is an arbitrary constant, then the value of $k$ is

1/2

-1/2

1/4

-1/4

✅ Correct Option: 4

$\tan x+\cot x = \frac{1}{\sin x\cos x}$ , so the integrand is $\sin x\cos x = \frac{1}{2}\sin 2x$ .

$\int\frac{1}{2}\sin 2x\,dx = -\frac{1}{4}\cos 2x+C$ , so $k = -\frac{1}{4}$ .

2025: 22 May Shift 2

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

2024: 16 May Shift 1

$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$

2026: 21 May Shift 1

$\int \frac{dx}{\sin^2 x \cos^2 x} =$

2026: 22 May Shift 1

$I=\int_{2}^{8}\frac{\sqrt[3]{x+1}}{\sqrt[3]{x+1}+\sqrt[3]{11-x}}dx$ is equal to :

2025: 3 June Shift 1

The value of $\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to (where C is an arbitrary constant)

2025: 22 May Shift 2

Match List-I with List-II (where $c$ is an arbitrary constant)

Choose the correct answer from the options given below: