_(()/6)^(()/3) dx/(1 + √( x)) is equal to : Option 1: ()/6 Option 2: ()/12 Option 3: ()/4 Option 4: ()/3

Answer: ()/12. Solution: Use property _a^b f(x)dx = _a^b f(a+b-x)dx with a+b = /2. Since √((/2 - x)) = 1/√( x), the second form of the integrand is √( x)/(1+√( x)). Adding gives 2I = _(/6)^(/3) 1 dx = /6, so I = /12.

Calculus

Integrals

Hard

core

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}$ is equal to :

$\frac{\pi}{6}$

$\frac{\pi}{12}$

$\frac{\pi}{4}$

$\frac{\pi}{3}$

✅ Correct Option: 2

Use property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ with $a+b = \pi/2$ . Since $\sqrt{\tan(\pi/2 - x)} = 1/\sqrt{\tan x}$ , the second form of the integrand is $\sqrt{\tan x}/(1+\sqrt{\tan x})$ . Adding gives $2I = \int_{\pi/6}^{\pi/3} 1\,dx = \pi/6$ , so $I = \pi/12$ .

2026: 18 May Shift 1

The value of $\int \frac{dx}{1+\sqrt{x}}$ is

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The integral I = $\int \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{3\log_e x} - e^{2\log_e x}} dx$ is equal to

2026: 19 May Shift 1

Match List-I with List-II

[c is an arbitrary constant]

List-I	List-II
(A) $\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}\,dx$	(I) $\tan x - x + c$
(B) $\int \frac{\sec^2 x}{\text{cosec}^2 x}\,dx$	(II) $2\tan x - 3\sec x + c$
(C) $\int \sec x(\sec x + \tan x)\,dx$	(III) $\tan x + \cot x + c$
(D) $\int \frac{2-3\sin x}{\cos^2 x}\,dx$	(IV) $\tan x + \sec x + c$

Choose the correct answer from the options given below:

2025: 27 May Shift 1

The value of $\int \frac{x^5}{\sqrt{1 + x^3}} dx$ is

2022: 23 Aug Shift 1

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

2025: 3 June Shift 2

$\int_0^8 (x^{\frac{2}{3}} + 1) dx$ is equal to

∫π6π3dx1+tan⁡x\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}∫6π​3π​​1+tanx​dx​ is equal to :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}$ is equal to :