The value of integral _(-/2)^(/2) 1/(1+e^( x))dx is Option 1: Option 2: ()/2 Option 3: ()/4 Option 4: 2

Answer: ()/2. Solution: Let I=_(-/2)^(/2)dx/(1+e^( x)). Replacing x by -x gives I=_(-/2)^(/2)(e^( x))/(1+e^( x))dx. Adding, 2I=_(-/2)^(/2)1 dx=, so I=()/2.

Calculus

Integrals

Medium

core

The value of integral $\int_{-\pi/2}^{\pi/2} \frac{1}{1+e^{\sin x}}dx$ is

$\pi$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$2\pi$

✅ Correct Option: 2

Let $I=\int_{-\pi/2}^{\pi/2}\frac{dx}{1+e^{\sin x}}$ . Replacing $x$ by $-x$ gives $I=\int_{-\pi/2}^{\pi/2}\frac{e^{\sin x}}{1+e^{\sin x}}dx$ .

Adding, $2I=\int_{-\pi/2}^{\pi/2}1\,dx=\pi$ , so $I=\frac{\pi}{2}$ .

2026: 21 May Shift 2

$\displaystyle\int \dfrac{x + \dfrac{a}{x}}{x^2 + \dfrac{a^2}{x^2}}\,dx, a > 0$ is equal to (where C is an arbitrary constant)

2026: 29 May Shift 1

$\displaystyle\int \dfrac{(x-3)e^x}{(x-1)^3}dx$ is equal to:

(where C is an arbitrary constant)

2025: 13 May Shift 2

$\int_0^2 x(2-x)^n dx$ is equal to

2022: 16 July Shift 2

The value of $\int_{-3}^{3} \log_e\left(\frac{4+x}{4-x}\right) dx$ is

2025: 22 May Shift 1

$\int \frac{dx}{2\sin^2 x + 5\cos^2 x}$ is equal to

2025: 29 May Shift 2

Match List-I with List-II

List-I	List-II
Integral	Solution: C is an arbitrary constant
---	---
(A) $\int \frac{dx}{x^2 + 25}$	(I) $\frac{1}{10} \log \left\vert \frac{5 + x}{5 - x} \right\vert + C$
(B) $\int \frac{dx}{x^2 - 25}$	(II) $\log \vert x + \sqrt{x^2 - 25}\vert + C$
(C) $\int \frac{dx}{25 - x^2}$	(III) $\frac{1}{5} \tan^{-1} \left( \frac{x}{5} \right) + C$
(D) $\int \frac{dx}{\sqrt{x^2 - 25}}$	(IV) $\frac{1}{10} \log \left\vert \frac{x - 5}{x + 5} \right\vert + C$

Choose the correct answer from the options given below:

The value of integral ∫−π/2π/211+esin⁡xdx\int_{-\pi/2}^{\pi/2} \frac{1}{1+e^{\sin x}}dx∫−π/2π/2​1+esinx1​dx is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The value of integral $\int_{-\pi/2}^{\pi/2} \frac{1}{1+e^{\sin x}}dx$ is