Question

integral of (x^2)/(7-x^2)^(3/2)

Answer 1

\[\int\limits (x^2)/(7-x^2)^{3/2}dx\]
Can you think of a trig substitution suitable for use where you have a denom. that looks like a^2-x^2?

Answer 2

Please note that \[(7-x^2)^{3/2}=(7-x^2)^{1/2}(7-x^2)\].

Answer 3

W\[\sqrt{a^2-x^2} \]here have you seen

Answer 4

before?

Answer 5

tan^2(T) = Sec^2(T) - 1

Answer 6

\[I=-\int\limits \frac{ -x^2 }{ \left( 7-x^2 \right)\sqrt{7-x^2} }dx=-\int\limits \frac{ 7-x^2-7 }{ \left( 7-x^2 \right)\sqrt{7-x^2} }dx+c\]
\[=-\int\limits \frac{ dx }{ \sqrt{7-x^2} }+7 \int\limits \frac{ dx }{ \left( 7-x^2 \right)\sqrt{7-x^2} }dx+c\]
\[=-?+7I _{1}+c\]
\[I _{1}=\int\limits \frac{ dx }{ \left( 7-x^2 \right)\sqrt{7-x^2} }\]
\[put~x=\frac{ 1 }{ y },dx=-\frac{ 1 }{ y^2 }dy\]
\[I _{1}=\int\limits \frac{y^2* \frac{ -1 }{ y^2 }*y~dy }{ \left( 7y^2-1 \right)\sqrt{7y^2-1} }=-\int\limits \left( 7y^2-1 \right)^{-\frac{ 3 }{ 2 }}y~dy \]=?