How do you integrate #int 1/sqrt(4x^2+16x) # using trigonometric substitution?

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Answer 1 · 2016-08-22T04:43:52

Let #I=int1/sqrt(4x^2+16x)dx=1/2int1/sqrt(x^2+4x)dx#

Now, #sqrt(x^2+4x)=sqrt(x^2+4x+4-4)=sqrt{(x+2)^2-2^2}#.

We take substn. #"(x+2)=2secy, so, dx=2secytanydy"#

Also, #sqrt(x^2+4x)=sqrt{4sec^2y-4}=sqrt(4tan^2y)=2tany#.

Hence, #I=1/2int(2secytany)/(2tany) dy=1/2intsecydy#

#=1/2ln|secy+tany|#

#=1/2ln|(x+2)/2+sqrt(x^2+4x)/2|+C#

#=1/2ln|(x+2+sqrt(x^2+4x))/2|+C#

#=1/2ln|x+2+sqrt(x^2+4x)|+C-1/2ln2#

#=1/2ln|x+2+sqrt(x^2+4x)|+K, "where", K=C-1/2ln2#.

Enjoy Maths.!