Question

intergal of (x^2)/(1-9x^2)^(3/2)) dx

Answer 1

\[\int\limits_{?}^{?}\frac{ x^2 }{ (\sqrt{1-9x^2})^3 } dx\]

Answer 2

should i use this \[\int\limits_{?}^{?}\frac{ 1 }{ \sqrt{1-x^2} }dx =\sin ^{-1}x + C\]

Answer 3

\[I=\int\limits \frac{ x^2 }{( \sqrt{1-9x^2})^3 }dx=-\frac{ 1 }{ 9 }\int\limits \frac{ -9x^2 }{ \left( 1-9x^2 \right) \sqrt{1-9x^2}} dx\]
\[=-\frac{ 1 }{ 9 }\int\limits \frac{ 1-9x^2-1 }{ \left( 1-9x^2 \right) \sqrt{1-9x^2}}dx\]
\[=-\frac{ 1 }{ 9 }\left[ \int\limits \frac{ 1 }{ \sqrt{1-9x^2} } dx-\int\limits \frac{ 1 }{ \left( 1-9x^2 \right)\sqrt{1-9x^2} }dx\right]+c\]
\[=-\frac{ 1 }{ 9 }\left[ I _{1}-I _{2} \right]\]

Answer 4

@surjithayer im sorry where did 1/9 come from can u go slowly step by step

Answer 5

just to make -9x^2, i multiply and divide by 9

Answer 6

then add and subtract 1 to make 1-9x^2

Answer 7

for I2 put 3x=sin theta
\[3 dx= \cos \theta d \theta \]
then complete i am leaving.

Answer 8

isnt it 3x=9sin theta?

Answer 9

\[dx=\frac{ 1 }{ 3 }\cos \theta d \theta\]
\[I _{2}=\int\limits \frac{ 1 }{ \left( 1-\sin ^2\theta \right) \sqrt{1-\sin ^2\theta }}\frac{ 1 }{ 3 }\cos \theta d \theta \]
\[=\frac{ 1 }{ 3 }\int\limits \frac{ \cos \theta d \theta }{ \cos ^2\theta \cos \theta }=\frac{ 1 }{ 3 }\int\limits \sec ^2\theta d \theta =\frac{ 1 }{ 3 }\tan \theta \]