Mathematics

If $\displaystyle I=\int \frac{\sin x\left ( \cos x \right )^{-5/2}dx}{\sqrt{\sin x+3\cos x}+\sqrt{\sin x+4\cos x}}$$\displaystyle= \frac { A }{ 1550 } \left( \left( \tan x+4 \right) ^{ 5/2 }-\left( \tan x+3 \right) ^{ 3/2 } \right) -\frac { 2 }{ 3 } \left[ 4\left( \tan x+4 \right) ^{ 3/2 }-3{ \left( \tan { x } +3 \right) }^{ 3/2 } \right] +C$,then the value of A is equal to

620

SOLUTION
$I=\int { \cfrac { \sin x\left( \cos x \right) ^{ -5/2 }dx }{ \sqrt { \sin x+3\cos x } +\sqrt { \sin x+4\cos x } } } =\int { \cfrac { \sin { x } \left( \cos { x } \right) ^{ -3 } }{ \sqrt { \tan { x } +3 } +\sqrt { \tan { x } +4 } } }$
Let $\tan { x } =t$
$I=\int { \cfrac { t }{ \sqrt { t+3 } +\sqrt { t+4 } } dt } ={ I }_{ 4 }-{ I }_{ 3 }$
Where ${ I }_{ a }=\int { t\sqrt { t+a } } dt$
Susbtituting $t+a={ u }^{ 2 }$
${ I }_{ a }=\int { \left( { u }^{ 2 }-a \right) u\left( 2u \right) du } =\cfrac { 2{ u }^{ 5 } }{ 5 } -\cfrac { 2a{ u }^{ 3 } }{ 3 } =\cfrac { 2{ \left( t+a \right) }^{ 5/2 } }{ 5 } -\cfrac { 2a{ \left( t+a \right) }^{ 3/2 } }{ 3 }$
Therefore
$I=\cfrac { 2 }{ 5 } \left( { \left( \tan { x } +4 \right) }^{ 5/2 }-{ \left( \tan { x } +3 \right) }^{ 5/2 } \right) -\cfrac { 2 }{ 3 } \left( 4{ \left( \tan { x } +4 \right) }^{ 3/2 }-3{ \left( \tan { x } +3 \right) }^{ 3/2 } \right) +c$
Therefore, $A=620$

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One Word Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

Realted Questions

Q1 Single Correct Hard
$\int\, \displaystyle \frac {x^3\, +\, 3x\, +\,2}{(x^2\, +\, 1)^2\, (x\, +\, 1)}\, dx=$
• A. $\displaystyle \frac { 1 }{ 2 } \log { \left( { x }^{ 2 }-1 \right) } +\frac { 1 }{ 2 } \tan ^{ -1 }{ x } +\frac { x }{ 1\, +\, x^{ 2 } } -\frac { 1 }{ 2 } \log { \left( 1+x \right) }$
• B. $\displaystyle \frac {3}{2}\, tan^{-1}\, x\, +\, \displaystyle \frac {1}{2}ln(1\, -\, x)\, +\, \displaystyle \frac {1}{4}\, ln (1\, -\, x^2)\, +\, \displaystyle \frac {x}{1\, -\, x^2}\, +\, c$
• C. $\displaystyle \frac {3}{2}\, tan^{-1}\, x\, +\, \displaystyle \frac {1}{2}ln(1\, +\, x)\, +\, \displaystyle \frac {1}{4}\, ln (1\, +\, x^2)\, +\, \displaystyle \frac {x}{1\, +\, x^2}\, +\, c$
• D. $\displaystyle \frac { 1 }{ 2 } \log { \left( { x }^{ 2 }+1 \right) } +\frac { 1 }{ 2 } \tan ^{ -1 }{ x } +\frac { x }{ 1\, +\, x^{ 2 } } -\frac { 1 }{ 2 } \log { \left( 1+x \right) }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int _{ 0 }^{ \pi }{ x } In\left( \sin { x } \right) dx=$
• A. $\dfrac { \pi }{ 2 } ln\ 2$
• B. $-\dfrac { \pi }{ 2 } ln2$
• C. $\ ln\ 2$
• D. $\dfrac { -{ \pi }^{ 2 } }{ 2 } ln\ 2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}dx$ is equal to
• A. $\displaystyle \log|\cos x+\sin x+1|+2x+5\log|1+\tan\frac{x}{2}|+c$
• B. $\displaystyle \log|\cos x+\sin x+1|+2x-5\log|1+\tan\frac{x}{2}|+c$
• C. $-\log|\cos x +\sin_{X} +1|+2x -5\displaystyle \log|1+\tan\frac{x}{2}|+c$
• D. $-\log|\cos x+\sin x +1|+2x +5\displaystyle \log|1+\tan\frac{x}{2}|+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle\int{\frac{x^2(1-\ln{x})}{\ln^4{x}-x^4}dx}$ is equal to
• A. $\displaystyle\frac{1}{2}\ln{\left(\frac{x}{\ln{x}}\right)}-\frac{1}{4}\ln{(\ln^2{x}-x^2)}+C$
• B. $\displaystyle\frac{1}{4}\ln{\left(\frac{\ln(x)+x}{\ln{x}-x}\right)}-\frac{1}{2}\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}+C$
• C. $\displaystyle\frac{1}{4}\left(\ln{\left(\frac{\ln(x)-x}{\ln{x}+x}\right)}+\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}\right)+C$
• D. $\displaystyle\frac{1}{4}\ln{\left(\frac{\ln(x)-x}{\ln{x}+x}\right)}-\frac{1}{2}\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}+C$

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