How do you integrate #int dx/(5-4x-x^2)^(5/2)# using trig substitutions?

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Answer 1 · 2018-05-01T10:30:11

# I=((x+2)(19-8x-2x^2))/(243sqrt(5-4x-x^2))+C#.

Let us note that,

#5-4x-x^2=9-(4+4x+x^2)=3^2-(x+2)^2#.

So, if we subst. #(x+2)=3siny," then, "dx=3cosydy#.

#:. I=intdx/(5-4x-x^2)^(5/2)#,

#=int(3cosydy)/(9-9sin^2y)^(5/2)#,

#=3/3^5int(cosydy)/cos^5y#,

#=1/81intdy/cos^4y#,

#=1/81intsec^2y*sec^2ydy#,

#=1/81int(tan^2y+1)sec^2ydy#.

Now we subst. #tany=t. :. sec^2ydy=dt.#

#:. I=1/81int(t^2+1)dt#,

#=1/81(t^3/3+t)#,

#=t/243(t^2+3)#.

Since, #t=tany#, we get,

#I=tany/243(tan^2y+3)#.

Now, #(x+2)/3=siny#.

#:. cosy=sqrt{1-(x+2)^2/9}=1/3sqrt(5-4x-x^2)#.

#:. tany=(x+2)/sqrt(5-4x-x^2)#.

Finally, #I=1/243*(x+2)/sqrt(5-4x-x^2){((x+2)/sqrt(5-4x-x^2))^2+3}, i.e., #

# I=((x+2)(19-8x-2x^2))/(243sqrt(5-4x-x^2))+C#.