How do you integrate int sqrt(5+4x-x^2) by trigonometric substitution?

1 Answer
Nov 25, 2016

The answer is =9/2arcsin((x-2)/3)+(x-2)/2sqrt(9-(x-2)^2)+C

Explanation:

Let's complete the square

5+4x-x^2=9-(2-x)^2

Let x-2=3sinu

dx=3cosudu

9-(2-x)^2=9-9sin^2u=9cos^2u

intsqrt(5+4x-x^2)dx=int3cosu*3cosudu

=9intcos^2udu

cos2u=2cos^2u-1

cos^2u=(1+cos2u)/2

9intcos^2udu=9/2int(1+cos2u)du

=9/2(u+(sin2u)/2)=9/2(u+sinucosu)

sinu=(x-2)/3

u=arcsin((x-2)/3)

cosu=1/3sqrt(9-(x-2)^2)

9/2(u+sinucosu)=9/2(arcsin((x-2)/3)+(x-2)/9sqrt(9-(x-2)^2))

intsqrt(5+4x-x^2)dx=9/2arcsin((x-2)/3)+(x-2)/2sqrt(9-(x-2)^2)+C