# What is the definite integral of y = square root(1-x^2) if x=0 to x=1.

### 1 Answer | Add Yours

I'll suggest to replace the variable x by the function sin t.

x = sin t

We'll differentiate both sides and we'll get:

dx = cos t dt

We'll change the limits of integration, too:

x = 0 => t = 0

x = 1 => t = pi/2

The definite integral will be evaluated using the Leibniz Newton formula:

Int sqrt[1 - (sin t)^2] cos t dt = F(pi/2) - F(0)

But, from Pythagorean identity, we'll have:

1 - (sin t)^2 = (cos t)^2

We'll take square root both sides:

sqrt [1 - (sin t)^2] = sqrt (cos t)^2

sqrt [1 - (sin t)^2] = cos t

Int sqrt[1 - (sin t)^2] cos t dt = Int (cos t)^2 dt

Int (cos t)^2 = Int (1+cos 2t)dt/2

Int (1+cos 2t)dt/2 = Int dt/2 + (1/2)*Int cos 2t dt

Int (1+cos 2t)dt/2 = t/2 + sin 2t/4

F(pi/2) - F(0) = pi/4 + sin pi/4 - 0 + sin 0/4

sin pi = 0

F(pi/2) - F(0) = pi/4

**The definite integral of the given function is: Int sqrt(1 - x^2) dx = pi/4**