Question

How do you find the integral of `1/sqrt(1-4x^2)dx`?

Answer 1

We can evaluate this integral by using a trigonometric substitution.

Let `sintheta =2x` Then

`x=1/2sintheta`

differentiate

`dx=1/2costheta d theta`

Make substitution

`1/2intcostheta/(sqrt(1-4(1/2sintheta)^2))d theta`

`1/2intcostheta/(sqrt(1-4(1/4sin^2theta)))d theta`

`1/2intcostheta/(sqrt(1-sin^2theta))d theta`

`1/2intcostheta/(sqrt(cos^2theta))d theta`

`1/2intcostheta/(costheta)d theta`

`1/2intd theta`

Integrating we have `1/2theta`

Defining `theta` in terms of `x`

`theta=arcsin(2x)` Therefore

`1/2arcsin(2x)+C`

Answer 2

The answer is: `1/2arcsin2x+c`.

Remembering that:

`int(f'(x))/sqrt(1-[f(x)]^2)dx=arcsinf(x)+c`,

so:

`int1/sqrt(1-4x^2)dx=int1/sqrt(1-(2x)^2)dx=1/2int2/(sqrt(1-(2x)^2))dx=`

`=1/2arcsin2x+c`.