√(1 - 49x^2) dx is equal to Option 1: x/2(√(1 - 49x^2)) + 1/98^(-1)7x + C Option 2: 7x/2√(1 + 49x^2) + 1/49^(-1)x + C Option 3: x/2√(1 + 1/7x^2) - 1/49^(-1)7x + C Option 4: x/2√(1 - 49x^2) + 1/14^(-1)7x + C

Answer: x/2√(1 - 49x^2) + 1/14^(-1)7x + C. Solution: Use the standard result √(a^2 - u^2) du = u/2√(a^2-u^2) + a^2/2^(-1)u/a + C. Here √(1-49x^2) = √(1-(7x)^2). Let u=7x, du = 7dx: √(1-49x^2) dx = x/2√(1-49x^2) + 1/14^(-1)(7x) + C.

Calculus

Integrals

Medium

$\int \sqrt{1 - 49x^2} \, dx$ is equal to

$\frac{x}{2}\left(\sqrt{1 - 49x^2}\right) + \frac{1}{98}\sin^{-1}7x + C$

$\frac{7x}{2}\sqrt{1 + 49x^2} + \frac{1}{49}\sin^{-1}x + C$

$\frac{x}{2}\sqrt{1 + \frac{1}{7x^2}} - \frac{1}{49}\sin^{-1}7x + C$

$\frac{x}{2}\sqrt{1 - 49x^2} + \frac{1}{14}\sin^{-1}7x + C$

✅ Correct Option: 4

Use the standard result $\int \sqrt{a^2 - u^2}\,du = \frac{u}{2}\sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C$ .

Here $\sqrt{1-49x^2} = \sqrt{1-(7x)^2}$ . Let $u=7x$ , $du = 7dx$ :

$\int \sqrt{1-49x^2}\,dx = \frac{x}{2}\sqrt{1-49x^2} + \frac{1}{14}\sin^{-1}(7x) + C$ .

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The value of $\int_{-a}^a f(x)dx$ where $f(x) = \frac{7^x}{1+7^x}$ is:

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

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Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{\pi/6}^{\pi/3} \cos x \, dx$	(I) $0$
(B) $\displaystyle\int_{-\pi}^{\pi} (\sin^{61}x + x^{123}) \, dx$	(II) $4$
(C) $\displaystyle\int_{0}^{\pi/2} \dfrac{1}{1+\tan^3 x} \, dx$	(III) $\dfrac{\sqrt{3}-1}{2}$
(D) $\displaystyle\int_{-2}^{2} (2 - \left\vert x \right\vert) \, dx$	(IV) $\dfrac{\pi}{4}$

Choose the correct answer from the options given below:

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:

2026: 25 May Shift 1

$\displaystyle\int_{-2}^{2} \dfrac{x^2}{1+5^x}\, dx$ is equal to:

∫1−49x2 dx\int \sqrt{1 - 49x^2} \, dx∫1−49x2​dx is equal to

Solution

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$\int \sqrt{1 - 49x^2} \, dx$ is equal to