Evaluate: `intsqrt((3-x)/(3+x))sin^(-1)(1/6sqrt(3-x))dx`

Explanation

Here, `I=intsqrt((3-x)/(3+x))"sin"^(-1)((1)/(sqrt(6))sqrt(3-x))dx`
Put `x=3cos 2 theta " and " dx= -6 sin 2 theta d theta`
`I=int sqrt((3-3 cos 2 theta)/(3+cos 2 theta))"sin"^(-1)((1)/(sqrt(6))sqrt(3-3 cos 2 theta))xx(-6 sin 2 theta)d theta`
`=int(sin theta)/(cos theta)"sin"^(-1)(sin theta)(-6 sin 2 theta) d theta`
`= -6 int theta(2sin^(2)theta)d theta= -6 int theta(1-cos 2theta) d theta`
`= -6{(theta^(2))/(2)-int theta cos 2 theta d theta}`
`= -6 {(theta^(2))/(2)-(theta (sin 2 theta)/(2)-int 1 ((sin2 theta)/(2))d theta)}`
` = -3 theta^(2)+6{(theta sin 2 theta)/(2)+(cos 2 theta)/(4)}+c`
`=(1)/(4){-3("cos"^(-1)((x)/(3)))^(2)+2 sqrt(9-x^(2))"cos"^(-1)((x)/(3))+2x}+c`