Question

How do you integrate `int x^2sqrt(16-x^2)` using trig substitutions?

Answer 1

`32arcsin(x/4)-8sin(4arcsin(x/4))+C`

Explanation:

Solving for:

`I=intx^2sqrt(16-x^2)dx`

Make the substitution `x=4sintheta` where `dx=4costhetad theta`:

`I=int16sin^2thetasqrt(16-16sin^2theta)4costhetad theta`

Bringing all the multiplicative constants out, including the `sqrt16=4`:

`I=256intsin^2thetacosthetasqrt(1-sin^2theta)d theta`

Since `sin^2theta+cos^2theta=1`, we see that `costheta=sqrt(1-sin^2theta)`:

`I=256intsin^2thetacos^2thetad theta`

Here, since `sin2theta=2sinthetacostheta`, we see that `sin^2 2theta=4sin^2thetacos^2theta`:

`I=64int4sin^2thetacos^2thetad theta`

`I=64intsin^2 2theta d theta`

Note that, according to the cosine double angle formula, `cos4theta=1-2sin^2 2theta`, so `sin^2 2theta=(1-cos4theta)/2`:

`I=32int(1-cos4theta)d theta`

`I=32intd theta-32intcos4thetad theta`

Integrating both (substitution can be used for the second integral, or just see it for the reverse chain rule of `sin4theta` that it is):

`I=32theta-8int4cos4thetad theta`

`I=32theta-8sin4theta`

Since `x=4sintheta`, solving for `theta` gives `theta=arcsin(x/4)` and `4theta=4arcsin(x/4)`. Thus:

`I=32arcsin(x/4)-8sin(4arcsin(x/4))+C`