# What is integral `int_0^(pi/2) ` (1 - sin x)^(1/2)dx?

*print*Print*list*Cite

### 1 Answer

You need to evaluate the given definite integral, hence, I suggest you to convert the expression under the square root into a square, such that:

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

You may expand the right side to check if the result is identically to the left side expression, such that:

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

Using the Pythagorean trigonometric identity, yields:

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

Using the double angle identity, yields:

`2sin(x/2)*cos(x/2) = sin 2*(x/2) = sin x`

Replacing `1` for `sin^2(x/2) + cos^2(x/2)` and sin x for `2sin(x/2)*cos(x/2)` yields:

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

Hence, expanding the square `(cos(x/2) - sin(x/2))^2` yields exactly the original radicand, thus, you may replace `(cos(x/2) - sin(x/2))^2` for `1 - sin x` , such that:

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

You need to use the following property, such that:

`sqrt(x^2) = |x|`

Reasoning by analogy, yields:

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

Using the property of linearity of definite integral, yields:

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

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

Using the fundamental theorem of calculus, yields:

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

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

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

**Hence, evaluating the given definite integral, under the given conditions, yields ` int_0^(pi/2) sqrt(1 - sin x)dx = 2sqrt2 - 2.` **

**Sources:**