Question

How do you integrate `int sqrt(1-x^2)` by trigonometric substitution?

Answer 1

`int sqrt(1-x^2)dx = (sin^-1x + xsqrt(1-x^2))/2 + C`

Explanation:

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

The way to approach these types of problems is try and make a comparison with a known trig identity. In the case we will use:

`sin^2A + cos^2A -=1 => cos^2A=1-sin^2A`

If you compare to the integrand, we make the following substitution:

Let `x=sintheta => 1-sin^2theta=cos^2theta`
`:. 1-x^2 = cos^2theta`
`:. sqrt(1-x^2) = costheta`

And `dx/(d theta) = costheta`

Substituting into the integral we have:
`I = int cos theta cos theta d theta` =
`:. I = int cos^2 theta d theta`

Now `cos2A-=cos^2A-sin^2A = 2cos^2A-1`
`:. cos^2A=1/2( 1+cos2A)`

And so;
`I = int 1/2( 1+cos2theta) d theta`
`2I = int (1+cos2theta) d theta`
`2I = theta + (sin2theta)/2 + C'`

We now need to use the identity `sin2A-=2sinAcosA`
`2I = theta + sinthetacostheta + C'`
`:. 2I = sin^-1x + xsqrt(1-x^2) + C'`
`:. I = (sin^-1x + xsqrt(1-x^2) + C')/2`
`:. I = (sin^-1x + xsqrt(1-x^2))/2 + C`