How do you integrate #int 1/(x^2+4)^(3/2)# by trigonometric substitution?

Rewrite with the rational exponent converted to a root:

#intdx/(sqrt((x^2+4)^3)#

Let #x=2tantheta#
#dx=2sec^2thetad theta#

Rewrite:

#int(2sec^2theta)/(sqrt((4tan^2theta+4)^3))d theta#

#int(2sec^2theta)/(sqrt((4(tan^2theta+1))^3))d theta#

Recall the identity

#tan^2theta+1=sec^2theta#, and apply it:

#int(2sec^2theta)/(sqrt(64(sec^2theta)^3))d theta#

#1/4intsec^2theta/sqrt(sec^6theta)d theta#

#1/4intsec^2theta/sec^3thetad theta#

#1/4intcosthetad theta=1/4sintheta+C#

We want to rewrite in terms of #x.# Recalling that #x=2tantheta, tantheta=x/2.# If #tantheta=x/2, sintheta=x/sqrt(x^2+4)#.

This could be deduced by drawing a right triangle with the sides opposite and adjacent to angle #theta# being labeled #x, 2# (respectively), making the hypotenuse #sqrt(x^2+4)#, and #sintheta=(opposite)/(hypoten use)=x/sqrt(x^2+4)#

So,

#intdx/(x^2+4)^(3/2)=1/4x/sqrt(x^2+4)+C#

Here,

#I=int1/(x^2+4)^(3/2)dx#

Let, #x=2tanu=>dx=2sec^2udu#

#andx^2+4=4tan^2u+4=4(tan^2u+1)=4sec^2u#

So,

#I=int(2sec^2u)/((4sec^2u)^(3/2))du=int(2sec^2u)/(2secu)^3du#

#=>I=int(2sec^2u)/(8sec^3u)du#

#=>I=int1/(4secu)du#

#=>I=1/4intcosdu#

#=>I=1/4sinu+c#

#=>I=1/2(sinu/cosu)cosu+c#

#=1/4tanu/secu+c#

#=1/4tanu/sqrt(tan^2u+1)+c#

#=1/4(x/2)/sqrt((x^2/4)+1)+c#

#=1/4*x/sqrt(x^2+4)+c#