Question

How do you find`int (x^2) / (sqrt(81 - x^2)) dx` using trigonometric substitution?

Answer 1

`=81/2sin^(-1)(x/9)-(xsqrt(81-x^2))/2+C`

Explanation:

Let `x=9sintheta`
Then `dx=9costheta d theta` and `sintheta=x/9`

Then the original integral becomes

`9int(81sin^2theta)/(sqrt(81-81sin^2theta))costheta d theta`

Now using the trig identities `sin^2theta+cos^2theta = 1` and
`sin2theta=2sinthetacostheta`, together with factorizations and laws of surds, we may compute the integral as

`9xx81int(sin^2theta)/(9sqrt(1-sin^2theta))costheta d theta`

`=81int(sin^2theta)/costheta costheta d theta`

`=81[theta/2-(sin2theta)/4]+C`

`=81/2sin^(-1)(x/9)-81/2*x/9*sqrt(81-x^2)/9+C`