Mathematics

# Evaluate :$\int { [\sqrt { tanx } +\sqrt { cotx } ] } dx$.

##### SOLUTION
$I= \displaystyle\int (\sqrt{\tan x}+\sqrt{\cot x})dx$

$\therefore I= \displaystyle\int \sqrt{\dfrac{\sin x}{\cos x}}+\sqrt{\dfrac{\cos x}{\sin x}}dx$

$\therefore I= \displaystyle\int \dfrac{\sin x+ \cos x}{\sqrt{\sin x \cos x}}dx$
$\sin x- \cos x =t$

$\therefore (\cos x+ \sin x)dx=dt$

$\therefore 1-2 \sin x \cos x =t^{2}$

$\therefore I= \displaystyle\int \dfrac{\sqrt{2}}{\sqrt{1-t^{2}}}dt$

$\therefore I=\sqrt{2}\sin^{-1}t+c$

$\therefore I=\sqrt{2} \sin^{-1}(\sin x- \cos x)+c$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
The value of the integral $\displaystyle \int^{1/\sqrt{3}}_{1/\sqrt{3}} \dfrac{x^4}{1-x^4} cos ^{-1} \dfrac{2x}{1+x^2} dx$
• A. $\dfrac{\pi}{\sqrt{3}} + log \dfrac{\sqrt{3} +1}{\sqrt{3} -1}$
• B. $\dfrac{1}{\sqrt{3}} + \dfrac{\pi}{2} log \dfrac{\sqrt{3} +1}{\sqrt{3} -1}$
• C. none of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\underset { x\rightarrow 0 }{ lim } \frac { xtan2x-2xtanx }{ { (1-cos2x) }^{ 2 } }$ equals :
• A. $\frac { 1 }{ 4 }$
• B. 1
• C. $-\frac { 1 }{ 4 }$
• D. $\frac { 1 }{ 2}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Integrate with respect to $x$:
$\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate: $\displaystyle \int (1+x -x^{-1})e^{x+x^{-1}}dx$
• A. $(x +1)e^{x+x^{-1}}+c$
• B. $(x -1)e^{x+x^{-1}}+c$
• C. $-xe^{x+x^{-1}}+c$
• D. $xe^{x+x^{-1}} + c$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let $\displaystyle 2I_{1}+I_{2}=\int \frac {e^{x}}{e^{2x}+e^{-2x}}dx$  and  $\displaystyle I_{1}+2I_{2}=\int \frac {e^{-x}}{e^{2x}+e^{-2x}}dx$
On the basis of above information, answer the following questions :