Question

How do you integrate `int x sqrt( 1 - x^4 )dx` using trigonometric substitution?

Answer 1

If we let `u = x^2`, then `du = 2xdx` and `dx= (du)/(2x)`.

`I = int xsqrt(1 - u^2)(du)/(2x)`

`I = 1/2intsqrt( 1 - u^2) du`

This is when the trig substitution comes in. Let `u = sintheta`. Then `du = costheta d theta`.

`I = 1/2int sqrt(1 - sin^2theta) costheta d theta`

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

`I = 1/2int costhetacostheta d theta`

`I = 1/2int cos^2theta d theta`

We know that `cos(2theta) = 2cos^2theta - 1`, so `1/2cos(2theta) = cos^2theta - 1/2` and `1/2cos(2theta) + 1/2 = cos^2theta`. This is the famous power reduction identity, frequently rewritten as `1/2(cos2theta + 1)`

`I = 1/2int 1/2(cos2theta + 1) d theta`

`I = 1/4int cos(2theta) + 1 d theta`

`I = 1/4(1/2sin(2theta)) + 1/4theta + C`

`I = 1/8sin(2theta) + 1/4theta + C`

Recall that `sin(2theta) = 2sinthetacostheta`.

`I = 1/8(2)sinthetacostheta + 1/4theta + C`

`I = 1/4sinthetacostheta + 1/4theta + C`

From our initial `theta`-substitution, we know that `sintheta = u/1`. Thus `theta = arcsin(u)` and `costheta = sqrt(1- u^2)`

`I = 1/4usqrt(1 - u^2) + 1/4arcsin(u) + C`

`I = 1/4x^2sqrt(1 - x^4) + 1/4arcsin(x^2) + C`

Hopefully this helps!