# int sin(sqrt(theta)) / sqrt(theta) d theta Find or evaluate the integral

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...

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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` .

Approved by eNotes Editorial Team