# What is definite integral `int_0^(pi/2)` (1+cos x)^(1/2) dx?

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### 1 Answer

You need to use the following half angle trigonometric identity, such that:

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

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

`int_0^(pi/2) (1 + cos x)^(1/2) dx = int_0^(pi/2) (2cos^2(x/2))^(1/2) dx`

`int_0^(pi/2) (2cos^2(x/2))^(1/2) dx = int_0^(pi/2) sqrt2*cos(x/2) dx`

Taking out the constant yields:

`int_0^(pi/2) sqrt2*cos(x/2) dx = sqrt 2*int_0^(pi/2) cos(x/2) dx`

`sqrt 2*int_0^(pi/2) cos(x/2) dx = sqrt2*(sin (x/2))/(1/2)|_0^(pi/2)`

By fundamental theorem of calculus, yields:

`sqrt 2*int_0^(pi/2) cos(x/2) dx = 2sqrt2*(sin (pi/4) - sin 0)`

`sqrt 2*int_0^(pi/2) cos(x/2) dx = 2sqrt2*(sqrt2/2) = 2`

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

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