How do you integrate #int 1/(sqrt(x^2+16))# by trigonometric substitution?

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Answer 1 · 2016-12-02T17:58:46

The answer is #=ln(abs(sqrt(16+x^2)/4+x/4))+C#

We use #tan^2x+1=sec^2x#

Let #x=4tanu#

#dx=4sec^2udu#

#intdx/sqrt(x^2+16)=int(4sec^2udu)/sqrt(16tan^2u+16)#

#=int(4sec^2udu)/(4secu)#

#=intsecudu#

#=int(secu(tanu+secu)du)/(tanu+secu)#

#=int((secutanu+sec^2u)du)/(tanu+secu)#

Let #secu+tanu=w#

#dw=(secutanu+sec^2u)du#

So,

#int((secutanu+sec^2u)du)/(tanu+secu)#

#=int(dw)/w#

#=ln(w)#

#=ln(secu+tanu)#

#tanu=x/4#

#sec^2u=1+tan^2u=1+x^2/16#

#ln(secu+tanu)=ln(sqrt(16+x^2)/4+x/4)#

#intdx/sqrt(x^2+16)=ln(abs(sqrt(16+x^2)/4+x/4))+C#