Question

How do you integrate `int sqrt(9-x^2)dx` using trigonometric substitution?

Answer 1

`1/2*x*sqrt(9-x^2)+9/2*arcsin(1/3*x)+C`

Explanation:

Substituting
`x=3sin(t)`
`dx=3cos(t)dt`
so we get
`9intsqrt(1-sin^2(t))cos(t)dt`
and we get
`9intcos^2(t)dt`
With `1/2*(cos(2t)+1)=cos^2(t)`
we get
`9/2int(cos(2t)+1)dt`
This is
`9/2(sin(2t)/2+t)+C`
Backsubstitution gives
`1/2xsqrt(9-x^2)+9/2arcsin(1/3x)+C`