Question

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

Answer 1

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`