# I know how to calculate indefinite integral of sin x but i'm stuck with sin (x)^(1/2).

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### 1 Answer

To solve the indefinite integral of sin sqrt x, I'll suggest to change the variable x into t^2, such as:

sin (sqrt x) = sin sqrt t^2 = sin t

x = t^2

We'll differentiate both sides:

dx = 2tdt

We'll re-write the inetgral:

Int sin (sqrt x) dx = Int 2t*sin t dt

We'll solve the integral by parts.

Int udv = u*v - Int v du

u = 2t => du = 2dt

dv = sin t dt => v = -cos t

Int 2t*sin t dt = -2t*cos t + 2*Int cos t dt

Int 2t*sin t dt = -2t*cos t + 2*sin t + C

The indefinite integral of sin sqrt x is:

**Int sin sqrt x dx = 2sin (sqrt x) - 2*sqrt x*cos (sqrt x) + C**