# Calculate `int_0^(pi/2) sqrt(1-sin x) dx` with sin `x/2` and cos `x/2` ?

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You need to evaluate the definite integral `int_0^(pi/2)sqrt(1-sin x)dx` using the sine and cosine of half angle such that:

Notice that you may substitute `cos^2 (x/2) + sin^2 (x/2)` for 1 and you may write `sin x = sin 2(x/2) = 2sin (x/2)cos(x/2)` such that:

`1 - sin x = cos^2 (x/2) - 2sin(x/2)*cos(x/2) + sin^2(x/2)`

`1 - sin x = (cos (x/2) - sin (x/2))^2`

`sqrt(1 - sin x) = sqrt((cos (x/2) - sin (x/2))^2) = |cos (x/2) - sin (x/2)|`

Hence, evaluating the definite integral yields:

`int_0^(pi/2) sqrt(1-sin x)dx = int_0^(pi/2) (cos (x/2) - sin (x/2))dx`

`int_0^(pi/2) sqrt(1-sin x)dx = int_0^(pi/2) cos (x/2)dx - int_0^(pi/2) sin (x/2)dx`

`int_0^(pi/2) sqrt(1-sin x)dx = 2sin(x/2)|_0^(pi/2) + 2cos(x/2)|_0^(pi/2)`

`int_0^(pi/2) sqrt(1-sin x)dx = 2sin(pi/4) - 2sin 0 + 2cos (pi/4) - 2cos 0`

`int_0^(pi/2) sqrt(1-sin x)dx = 2sqrt2/2 - 0 + 2sqrt2/2 - 2`

`int_0^(pi/2) sqrt(1-sin x)dx = 2sqrt2 - 2`

**Hence, evaluating the definite integral using the sine and cosine of half angle yields int_0^(pi/2) sqrt(1-sin x)dx = 2sqrt2 - 2. **

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