Mathematics

# The value of $\displaystyle\int { \dfrac { \sqrt { \tan { x } } }{ \sin { x } \cos { x } } } dx$ is equal to

$2\sqrt { \tan { x } }+C$

##### SOLUTION
Given $\displaystyle\int \dfrac{\sqrt{\tan x}}{\sin x\cos x}$
simplifying the function
$=\displaystyle\int \dfrac{\sqrt{\tan x}}{\sin x.\cos x.\dfrac{\cos x}{\cos x}}$
$=\displaystyle\int\dfrac{\sqrt{\tan x}}{\sin x.\dfrac{\cos^{2}x}{\cos x}}$
$=\displaystyle\int\dfrac{\sqrt{\tan x}}{\cos^{2}x.\dfrac{\sin x}{\cos x}}$
$=\displaystyle\int\dfrac{\sqrt{\tan x}}{\cos^{2}x.\tan x}$
$=\displaystyle\int\dfrac{\sqrt{\tan x}.(\tan x)^{-1}}{\cos^{2}x}$
$=\displaystyle\int\dfrac{(\tan x)^{1/2^{-1}}}{\cos^{2}x}$
$=\displaystyle\int\dfrac{(\tan x)^{-1/2}}{\cos^{2}x}$
$=\displaystyle\int(\tan x)^{-1/2}.\dfrac{1}{\cos^{2}x}$
$=\displaystyle\int (\tan x)^{-1/2}.\sec^{2} x$
Let $\tan x=t$
So, $\sec^{2}x=\dfrac{dt}{dx}$
$\Rightarrow dx=\dfrac{dt}{\sec^{2}x}$
$\therefore \displaystyle\int (\tan x)^{-1/2}.\sec^{2}x.dx$
$=\displaystyle\int (t)^{-1/2}.\sec^{2}x.\dfrac{dt}{\sec^{2}x}$
$=\displaystyle\int (t)^{-1/2} dt$
$=\dfrac{t^{-1/2}+1}{-1/2 +1}+C$      $\left\{as \displaystyle\int x^{n}dx=\dfrac{x^{n+1}}{n+1}+C\right\}$
$=\dfrac{t^{1/2}}{1/2}+C$
$=2 t^{1/2}+C$
$=2\sqrt{t}+C$
Substituting $t=\tan x$
$=2\sqrt{\tan x}+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
$\int \sqrt { 1 - \sin x } d x =$
• A. $2\sqrt {1-\sin x}+C$
• B. $2\sqrt {1-2\sin x}+C$
• C. $2\sqrt {1-\sin 2x}+C$
• D. $2\sqrt {1+\sin x}+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium

$\displaystyle \int_{0}^{\pi/2}\frac{\cos ec^{2/3}x}{\cos ec^{2/3}x+\sec^{2/3}x}d_{X=}$
• A. $\pi$
• B. $-\pi$
• C. $0$
• D. $\pi/4$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate:
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1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\int x^{3}d(tan^{-1}x)$ is equal to

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$