Calculus

Integrals

Easy

$\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx =$

$-1$

$1$

$0$

$\pi$

✅ Correct Option: 3

$\sin^7 x$ is an odd function. The integral of an odd function over a symmetric interval $[-a, a]$ is $0$ .

2025: 29 May Shift 2

$\int (e^{x\log a} + e^{a\log x}) dx$ is equal to (where $a > 1$ )

2025: 14 May Shift 1

The integral I = $\int e^x\left(\frac{x - 1}{3x^2}\right) dx$ is equal to

2023: 15 June Shift 2

The integral $\int_{0}^{1} x(1-x)^n dx$ is equal to :

2026: 19 May Shift 1

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

2025: 27 May Shift 1

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

2026: 22 May Shift 1

Match List-I with List-II

Where C is arbitrary constant.

List-I	List-II
(A) $\displaystyle\int \dfrac{dx}{x^2 - a^2} =$	(I) $\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$
(B) $\displaystyle\int \dfrac{dx}{a^2 - x^2} =$	(II) $\log\left\vert x + \sqrt{x^2-a^2}\right\vert + C$
(C) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} =$	(III) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a}\right\vert + C$
(D) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} =$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{a+x}{a-x}\right\vert + C$

Choose the correct answer from the options given below:

∫−π/2π/2sin⁡7x dx=\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx =∫−π/2π/2​sin7xdx=