$\displaystyle\int\dfrac{\sqrt{\tan x}}{\sin 2x}\,dx$ is: (a) $\dfrac{1}{2}\sqrt{\tan x}+C$ (b) $\sqrt{\tan x}+C$ (c) $\dfrac{1}{4}\sqrt{\tan x}+C$ (d) $2\sqrt{\tan x}+C$

∫dfrac√(tan x)sin 2xdx is: (a) (1)/(2)√(tan x)+C (b) √(tan x)+C (c) (1)/(4)√(tan x)+C (d) 2√(tan x)+C

Question 121 of 129

\displaystyle\int\dfrac{\sqrt{\tan x}}{\sin 2x}\,dx

(a)

\dfrac{1}{2}\sqrt{\tan x}+C

(b)

\sqrt{\tan x}+C

(c)

\dfrac{1}{4}\sqrt{\tan x}+C

(d)

2\sqrt{\tan x}+C

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2024MCQ· 1mImportance★★★★★

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The integral equals $\sqrt{\tan x}+C$ .

Let $u=\tan x$ , so $du=\sec^2x\,dx$ , i.e. $dx=\cos^2x\,du$ .

Also $\sin2x=\dfrac{2\tan x}{1+\tan^2x}=\dfrac{2u}{1+u^2}$ and $\cos^2x=\dfrac{1}{1+u^2}$ .

So the integrand becomes

$\dfrac{\sqrt u}{\frac{2u}{1+u^2}}\cdot\dfrac{1}{1+u^2}\,du=\dfrac{\sqrt u(1+u^2)}{2u(1+u^2)}\,du=\dfrac{1}{2\sqrt u}\,du.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2025Set jan-20251 markMCQ
$\displaystyle\int (2x + 5)^7 \, dx$
›Reveal solutionSolution

$\int (2x+5)^7\,dx = \dfrac{(2x+5)^8}{2 \cdot 8} = \dfrac{(2x+5)^8}{16} + C$ .

Step 1 — Use the standard form for a linear substitution
$\int (ax+b)^n\, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$
Step 2 — Identify the constants
$a = 2, \quad b = 5, \quad n = 7$
Step 3 — Substitute
$\int (2x+5)^7\, dx = \frac{(2x+5)^{8}}{2 \times 8} = \frac{(2x+5)^8}{16} + C$
Why the other options are wrong: (B) divides only by 7 (raises to the wrong power AND drops the $\frac{1}{a}$ ); (C) and (D) wrongly expand $(2x+5)$ as $(2x^2+5x)$ , which is not the inner function. …
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CA Foundation 2024Set sep-20241 markMCQ
$\int \dfrac{1}{x(x^5 + 1)} \, dx$
›Reveal solutionSolution

Substitute u = x⁵ then use partial fractions: the integral = (1/5)·ln(x⁵/(x⁵+1)) + c.

Step 1 — Prepare for substitution

Multiply top and bottom by $x^4$ :
$\int \dfrac{1}{x(x^5+1)}\,dx = \int \dfrac{x^4}{x^5(x^5+1)}\,dx$
Step 2 — Substitute u = x⁵

With $u = x^5$ , $du = 5x^4\,dx$ , i.e. $x^4\,dx = \dfrac{du}{5}$ :
$= \dfrac{1}{5}\int \dfrac{du}{u(u+1)}$
Step 3 — Partial fractions and integrate
$\dfrac{1}{u(u+1)} = \dfrac{1}{u} - \dfrac{1}{u+1}$ $\dfrac{1}{5}\int\!\left(\dfrac{1}{u} - \dfrac{1}{u+1}\right)du = \dfrac{1}{5}\big[\ln u - \ln(u+1)\big] + c = \dfrac{1}{5}\ln\!\left(\dfrac{x^5}{x^5+1}\right) + c$
Why the other options are wrong: (A) omits the 1/5 factor; (C) uses 1/3 (wrong power); (D) inverts the ratio to (x⁵+1)/x⁵. Only (B) has both the 1/5 and the correct x⁵/(x⁵+1). …
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Q. $\displaystyle\int\dfrac{\sqrt{\tan x}}{\sin 2x}\,dx$ is:

Concept understanding — Integration by Substitution

Step 1 — Use the standard form for a linear substitution

Step 2 — Identify the constants

Step 3 — Substitute

Step 1 — Prepare for substitution

Step 2 — Substitute u = x⁵

Step 3 — Partial fractions and integrate

Q.∫tan⁡xsin⁡2x dx\displaystyle\int\dfrac{\sqrt{\tan x}}{\sin 2x}\,dx∫sin2xtanx​​dx is:

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Q. $\displaystyle\int\dfrac{\sqrt{\tan x}}{\sin 2x}\,dx$ is:

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