`int sin(theta)/(3-2cos(theta)) d theta` Find or evaluate the integral

`int ( sin (theta))/(3-2cos(theta)) d theta`

To solve, apply u-substitution method.

`u = 3-2cos (theta)`

`du = -2*(-sin (theta)) d theta`

`du = 2sin (theta) d theta`

`1/2du= sin (theta) d theta`

Expressing the integral in terms of u, it becomes

`= int 1/(3-2cos (theta)) * sin (theta) d theta`

`=int 1/u * 1/2 du`

`=1/2 int 1/u du`

Then, apply the integral formula `int 1/x dx = ln|x| + C` .

`= 1/2 ln|u| + C`

And substitute back `u = 3-2cos(theta)` .

`=1/2ln|3-2cos(theta)| + C`

Therefore, `int ( sin (theta))/(3-2cos(theta)) d theta = 1/2 ln|3-2cos(theta)| + C` .

Comments

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