Correct Answer is : 2tan⁡x+c2 { x}+c2tanx+c, where ccc is a constant of integration

2 { x}+c, where c is a constant of integration

{2}{{ x}}+c, where c is a constant of integration

Correct Answer is : 2tan⁡x+c2 { x}+c2tanx+c, where ccc is a constant of integration

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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 147 of 150

Marks: +1, -0

\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x=

Solution:

Concept:

Simplify the integrand using trigonometric identities, then substitute

u = \tan x

Explanation:

Rewrite

\sin x \cos x

\tan x \cos^2 x

\frac{\sqrt{\tan x}}{\sin x \cos x} = \frac{\sqrt{\tan x}}{\tan x \cos^2 x} = \frac{\sec^2 x}{\sqrt{\tan x}}.

So the integral becomes:

I = \int \frac{\sec^2 x}{\sqrt{\tan x}}\, dx.

Let

u = \tan x

, then

du = \sec^2 x\, dx

Substitute and integrate:

I = \int \frac{du}{\sqrt{u}} = \int u^{-1/2}\, du = 2\sqrt{u} + c.

Replace

u

with

\tan x

\int \frac{\sqrt{\tan x}}{\sin x \cos x}\, dx = 2\sqrt{\tan x} + c.

Answer:

2 \sqrt{\tan x}+c

, where

c

is a constant of integration.

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