Question

How do you integrate `int x^2 / sqrt(16+x^2) dx` using trigonometric substitution?

Answer 1

`I = 1/2xsqrt(x^2 + 16) - ln|(x + sqrt(x^2 + 16))/2| + C`

Explanation:

Since the square root in the denominator is of the form `sqrt(a^2 + x^2)`, we will use the substitution `x = atantheta = 4tantheta`.

Therefore, `dx =4sec^2theta d theta`.

`I = int (4tantheta)^2/sqrt(16 + (4tantheta)^2) 4sec^2theta d theta`

`I = int (16tan^2theta)/sqrt(16(1 + tan^2theta)) 4sec^2theta d theta`

`I = int(16tan^2theta)/sqrt(16sec^2theta) 4sec^2theta`

`I = int(16tan^2theta)/(4sectheta) * 4sec^2theta`

`I = int16tan^2thetasectheta`

`I = int16(sec^2theta - 1)sectheta`

`I = 16int sec^3theta - sectheta`

These are two known integrals.

`I = 8secthetatantheta - 8ln|tantheta + sectheta| + C`

Now recall that `x/4 = tantheta`, therefore `sectheta = sqrt(x^2 + 16)/4`.

`I = 1/2xsqrt(x^2 + 16) - 8ln|(x + sqrt(x^2 + 16))/2| + C`

Hopefully this helps!