How do you integrate #int 1/sqrt(4x^2+4x-24)dx# using trigonometric substitution?

Question

1478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-30T20:05:54

#I=1/2ln|2x+1+sqrt(4x^2+4x-24)|+C#

Here,

#I=int1/sqrt(4x^2+4x-24)dx#

#=int1/sqrt(4x^2+4x+1-25)dx#

#=int1/sqrt((2x+1)^2-5^2)dx#

Substituting,

#2x+1=5secu=>2dx=5secutanudu#

#=>dx=5/2secutanudu and color(blue)(secu=(2x+1)/5#

So,

#I=int1/sqrt(5^2sec^2u-5^2)xx5/2secutanudu#

#=5/2int (secutanu)/(5tanu)du#

#=1/2intsecudu#

#=1/2ln|secu+tanu|+c#

#=1/2ln|secu+sqrt(sec^2u-1)|+c,where, color(blue)(secu=(2x+1)/5#

#=1/2ln|(2x+1)/5+sqrt((((2x+1)/5))^2-1) |+c#

#=1/2ln|(2x+1)/5+sqrt(4x^2+4x+1-25)/5|+c#

#=1/2ln|((2x+1)+sqrt(4x^2+4x-24))/5|+c#

#=1/2ln|2x+1+sqrt(4x^2+4x-24)|-1/2ln(5)+c#

#=1/2ln|2x+1+sqrt(4x^2+4x-24)|+C# , Put , #C=c-1/2ln(5)#