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

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

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