({ x} + { x}) dx = [AP EAPCET 24 May 2025 S1]

Correct Answer is : 2Tan−1(tan⁡x−12tan⁡x)+c{2} {Tan}^{-1} ≤ft( { x - 1}{{2 x}} ) + c2Tan−1(2tanxtanx−1)+c

2 {Tan}^{-1} ≤ft( { x - 1}{{ x}} ) + c

{Tan}^{-1} ≤ft( { x - 2}{2 { x}} ) + c

{2} {Tan}^{-1} ≤ft( { x - 1}{{2 x}} ) + c

{2} {Tan}^{-1} ≤ft( { x + 1}{{2} x} ) + c

Correct Answer is : 2Tan−1(tan⁡x−12tan⁡x)+c{2} {Tan}^{-1} ≤ft( { x - 1}{{2 x}} ) + c2Tan−1(2tanxtanx−1)+c

Test Index

Indefinite Integration

Section: Mathematics

Question : 4 of 89

Marks: +1, -0

\int (\sqrt{\tan x} + \sqrt{\cot x}) dx =

$2 \text{Tan}^{-1} \left( \frac{\tan x - 1}{\sqrt{\tan x}} \right) + c$
$\text{Tan}^{-1} \left( \frac{\tan x - 2}{2 \sqrt{\tan x}} \right) + c$
$\sqrt{2} \text{Tan}^{-1} \left( \frac{\tan x - 1}{\sqrt{2 \tan x}} \right) + c$
$\sqrt{2} \text{Tan}^{-1} \left( \frac{\tan x + 1}{\sqrt{2} \tan x} \right) + c$

Solution:

\; ( \sqrt{\tan x} + \sqrt{\cot x} ) dx

\; = \int \left( \sqrt{\tan x} + \; \frac{1}{\sqrt{\tan x}} \right) dx

\; = \int \; \frac{\tan x + 1}{\sqrt{\tan x}} dx

Let

u = \sqrt{\tan x}

, then

u^2 = \tan x

and

2 u du \; = \sec^2 x dx

\; = (1 + \tan^2 x) dx

\; = (1 + u^4) dx

\; = \int \; \frac{u^2 + 1}{u} \cdot \; \frac{2u}{1+u^4} du

\; = \int \; \frac{2 (u^2+1)}{1+u^4} du = \int \; \frac{2 \left(1 + \frac{1}{u^2}\right)}{\frac{1}{u^2} + u^2} du

\; = \int \; \frac{2 \left(1 + \frac{1}{u^2}\right)}{\left(u - \frac{1}{u}\right)^2 + 2} du

Again, Let

v = u - \frac{1}{u}

\Rightarrow \; \; dv \; = \left(1 + \; \frac{1}{u^2}\right) du

\; = \int \; \frac{2}{2 \left( \frac{v^2}{2} + 1 \right)} dv = \int \; \frac{1}{\left( \frac{v^2}{2} + 1 \right)} dv

\; = \int \; \frac{1}{\left( \frac{v}{\sqrt{2}} \right)^2 + 1} dv

\; = \tan^{-1} \left( \frac{v}{\sqrt{2}} \right) \cdot \sqrt{2} + C

where

c =

constant

\; = \sqrt{2} \tan^{-1} \left( \frac{u - \frac{1}{u}}{\sqrt{2}} \right) + C

\; = \sqrt{2} \tan^{-1} \left( \frac{ \sqrt{\tan x} - \frac{1}{\sqrt{\tan x}} }{\sqrt{2}} \right) + C

\; = \sqrt{2} \tan^{-1} \left( \frac{\tan x - 1}{\sqrt{2 \tan x}} \right) + C

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