# What is the indefinite integral of arcsin, with arcsin equal to the inverse function of sin(x) with x out of [-pi/2,+pi/2]?Then we can also verify : Integral (1/2 to 1) of arcsin(x) dx = 5*pi/12 -...

What is the indefinite integral of arcsin, with arcsin equal to the inverse function of sin(x) with x out of [-pi/2,+pi/2]?

Then we can also verify : Integral (1/2 to 1) of arcsin(x) dx = 5*pi/12 - sqrt(3)/2

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### 1 Answer

You need to remember the following formula of integral such that:

`int arcsin x dx = x arcsin x + sqrt(1-x^2) + c`

Hence, evaluating the definite integral of arcsin x yields:

`int_(1/2)^1 arcsin x dx = (x arcsin x + sqrt(1-x^2))|_(1/2)^1`

`int_(1/2)^1 arcsin x dx = arcsin 1 + sqrt(1-1) - (arcsin(1/2))/2 - sqrt(1-1/4)`

`` `int_(1/2)^1 arcsin x dx = pi/2 + 0 - pi/12 - (sqrt3)/2`

Bringing the fractions `pi/2; pi/12` to the same denominator yields:

`pi/2 - pi/12 = (6pi - pi)/12 =(5pi)/12`

`` `int_(1/2)^1 arcsin x dx = (5pi)/12 - (sqrt3)/2`

**Hence, evaluating`int_(1/2)^1 arcsin x dx ` using`int arcsin x dx = x arcsin x + sqrt(1-x^2) + c` yields `int_(1/2)^1 arcsin x dx = (5pi)/12 - (sqrt3)/2.` **