Question

How do you find the antiderivative of `int x^2sqrt(1-x^2) dx`?

Answer 1

`1/8sin^-1(x)+1/8x(2x^2-1)sqrt(1-x^2)+C`

Explanation:

`I=intx^2sqrt(1-x^2)dx`

Let `x=sin(theta)`. This implies that `dx=cos(theta)d theta`. Then:

`I=intsin^2(theta)sqrt(1-sin^2(theta))(cos(theta)d theta)`

`I=intsin^2(theta)cos^2(theta)d theta`

Now let `cos^2(theta)=1-sin^2(theta)`:

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

`I=intsin^2(theta)d theta-intsin^4(theta)d theta`

Now use the identity `sin^2(alpha)=1/2(1-cos(2alpha))`:

`I=1/2int(1-cos(2theta))d theta-int(1/2(1-cos(2theta)))^2d theta`

`I=1/2intd theta-1/2intcos(2theta)d theta-1/4int(1-2cos(2theta)+cos^2(2theta))d theta`

Use the identity `cos^2(alpha)=1/2(1+cos(2alpha))`:

`I=1/2intd theta-1/2intcos(2theta)d theta-1/4intd theta+1/2intcos(2theta)d theta-1/8int(1+cos(4theta))d theta`

`I=1/4intd theta-1/8intd theta-1/8intcos(4theta)d theta`

Solving through a substitution:

`I=1/8theta-1/32sin(4theta)`

We now have to return to `x` from `theta`. Recall that `x=sin(theta)`. Thus `theta=sin^-1(x)`. Also, use the identity `sin(2alpha)=2sin(alpha)cos(alpha)` to break down `sin(4theta)`:

`I=1/8sin^-1(x)-1/16sin(2theta)cos(2theta)`

Reusing the sine double-angle identity and also applying `cos(2alpha)=1-2sin^2(alpha)`:

`I=1/8sin^-1(x)-1/8sin(theta)cos(theta)(1-2sin^2(theta))`

Flipping the order of the terms in the last parenthesis with the preceding negative sign and using `cos(x)=sqrt(1-sin^2(x))`:

`I=1/8sin^-1(x)+1/8sin(theta)sqrt(1-sin^2(theta))(2sin^2(theta)-1)`

Finally using `x=sin(theta)`:

`I=1/8sin^-1(x)+1/8x(2x^2-1)sqrt(1-x^2)+C`