How do you integrate $int dx/(4x^2+9)^2$ using trig substitutions?

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Answer 1 · 2018-03-21T02:16:02

$int dx/(4x^2+9)^2=1/108arctan((2x)/3)+x/(72x^2+162)+C$

$int dx/(4x^2+9)^2$

= $1/2int (2dx)/((2x)^2+3^2)^2$

After using $2x=3tanu$ and $2dx=3(secu)^2*du$ transforms, this integral became

$1/2int (3(secu)^2*du)/(81(secu)^4)$

= $1/54int (cosu)^2*du$

= $1/108int (1+cos2u)*du$

= $1/108u+1/216sin2u+C$

= $1/108u+1/216*(2tanu)/((tanu)^2+1)+C$

After using $2x=3tanu$ , $tanu=(2x)/3$ and $u=arctan((2x)/3)$ inverse transforms, I found

$int dx/(4x^2+9)^2$

= $1/108arctan((2x)/3)+1/216*(2*(2x)/3)/(((2x)/3)^2+1)+C$

= $1/108arctan((2x)/3)+x/(72x^2+162)+C$

How do you integrate int dx/(4x^2+9)^2int dx/(4x^2+9)^2 using trig substitutions?