If 0 < x < ()/4, then √(1-2x) dx is: [where C is an arbitrary constant] Option 1: x+ x+C Option 2: x- x+C Option 3: x- x+C Option 4: x+ x+C

Answer: x+ x+C. Solution: 1-2x=( x- x)^2, so √(1-2x)=, x- x, . For 0 x, so the integrand is x- x. Integrating gives x+ x+C.

Calculus

Integrals

Medium

core

If $0 < x < \frac{\pi}{4}$ , then $\int\sqrt{1-\sin2x}\,dx$ is:

[where C is an arbitrary constant]

$\sin x+\cos x+C$

$\sin x-\cos x+C$

$-\sin x-\cos x+C$

$-\sin x+\cos x+C$

✅ Correct Option: 1

$1-\sin2x=(\sin x-\cos x)^2$ , so $\sqrt{1-\sin2x}=|\sin x-\cos x|$ .

For $0<x<\frac{\pi}{4}$ , $\cos x>\sin x$ , so the integrand is $\cos x-\sin x$ .

Integrating gives $\sin x+\cos x+C$ .

2026: 25 May Shift 1

If $\displaystyle\int \dfrac{\sin^4 x}{\cos^8 x}\, dx = a\tan^7 x + b\tan^5 x + C$ , where 'C' is an arbitrary constant then

2026: 19 May Shift 1

The value of $\int_{0}^{3} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx$ is

2026: 15 May Shift 1

If $\displaystyle\int_0^3 (8x^3+5x^2-2x-k)\,dx = 3$ , where $k$ is constant, then $k$ is equal to

2025: 3 June Shift 2

$\int e^{2x}(\sin x + \frac{1}{2}\cos x) dx$ is equal to

2025: 13 May Shift 1

$\int \frac{dx}{9x^2 - 16}$ is equal to

2025: 16 May Shift 1

Match List-I with List-II

List-I	List-II
Functions	Integrals
---	---
(A) $\int \frac{dx}{x^2 - 4}, x \neq \pm 2$	(I) $\log \vert x + \sqrt{4 + x^2}\vert + C$ , where $C$ is an arbitrary constant
(B) $\int \frac{1}{\sqrt{16 - x^2}} dx; \vert x\vert < 4$	(II) $\sin^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant
(C) $\int \frac{1}{16 + x^2} dx$	(III) $\frac{1}{4} \log \left\vert \frac{x-2}{x+2} \right\vert + C$ , where $C$ is an arbitrary constant
(D) $\int \frac{1}{\sqrt{4 + x^2}} dx$	(IV) $\frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant

Choose the correct answer from the options given below: