Mathematics

Evaluate $\displaystyle \int_{}^{} {x\sqrt {\frac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx }$

$\displaystyle \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \frac{1}{2}\sqrt {{a^4} - {x^4}} + C$

SOLUTION
Consider $i=\displaystyle \int{x\dfrac{\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}}}dx$

put $x^2=t$ $\Rightarrow xdx=\dfrac{dt}{2}$

$I=\dfrac{1}{2}\displaystyle \int{\dfrac{\sqrt{a^2-t}}{\sqrt{a^2+t}}}dt=\dfrac{1}{2}\displaystyle \int{\dfrac{a^2-t}{\sqrt{a^4-t^2}}}dt$

$I=\dfrac{1}{2}\displaystyle \int{\dfrac{a^2dt}{\sqrt{a^4-t^2}}}-\dfrac{1}{2}\displaystyle \int{\dfrac{t}{\sqrt{a^4-t^2}}}dt$

$=\dfrac{a^2}{2}\sin^{-1}\dfrac{t}{a^2}+\dfrac{1}{4}\displaystyle \int{\dfrac{-2tdt}{\sqrt{a^4-t^2}}}$ put $a^4-t^2=u\\-2tdt=du$

$=\dfrac{a^2}{2}\sin^{-1}\dfrac{x^2}{a^2}+\boxed{\dfrac{1}{4}\displaystyle \int{\dfrac{du}{\sqrt u}}}\rightarrow \dfrac{1}{2}\sqrt u+c$

$=\dfrac{a^2}{2}\sin^{-1}\dfrac{x^2}{a^2}+\dfrac{1}{2}\sqrt{a^4-x^4}+c$.

Its FREE, you're just one step away

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

Realted Questions

Q1 Single Correct Medium
$\displaystyle \int\frac{e^{x}(1+x)}{sin^{2}(xe^{x})}dx=$
• A. $\tan(xe^{x})+c$
• B. $\sin(xe^{x})+c$
• C. $\sin e^{x}+c$
• D. $-\cot(xe^{x})+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate $\int \left ( 3-2x \right )\sqrt{2+x-x^{2}dx}.$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Find the value of the equation :  $\int _ { \ln \lambda } ^ { \ln \left( \dfrac { 1 } { \lambda } \right) } \dfrac { f \left( \dfrac { x ^ { 2 } } { 3 } \right) ( f ( x ) + f ( - x ) ) } { g \left( 3 x ^ { 2 } \right) ( g ( x ) - g ( - x ) ) } d x =?$   where  $\lambda > 1$
• A. $1$
• B. $\lambda$
• C. $1 / \lambda$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate : $\displaystyle\int \dfrac{\cos^{-1}x}{x^2} dx$

$\displaystyle \int x.(\log x)^2 \ dx$