Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 59

int (sin sqrt theta)/sqrt theta d theta
To solve, apply u-substitution method.

u=sqrt theta
u= theta ^(1/2)
du = 1/2 theta^(-1/2) d theta
du = 1/(2theta^(1/2))d theta
du =1/(2 sqrt theta) d theta
2du =1/sqrt theta d theta

Expressing the integral in terms of u, it becomes:
= int sin (sqrt theta) * 1/sqrt theta d theta
= int sin (u) * 2du
= 2 int sin (u) du
Then, apply the integral formula int sin (x) dx = -cos(x) + C .
= 2*(-cos (u)) + C
= -2cos(u) + C
And, substitute back u = sqrt theta .
= -2cos( sqrt theta) + C

Therefore, int (sin sqrt theta)/sqrt theta d theta= -2cos( sqrt theta) + C .