Question

Can someone explain how this step was computed? (Trig identities)

Answer 1

I don't understand how you can go from
$$\frac{25}{2}\int (1-cos(2\theta)) \; d\theta$$
to

$$\frac{25}{2}\int (\theta-\frac{1}{2}sin(2\theta)) \; d\theta$$

Answer 2

\[ \int (1-cos(2\theta)) \; d\theta \\
\int d\theta - \cos(2\theta) d\theta\\
\theta - \int \cos(2\theta) d\theta
\]
I assume it makes sense that the integral of dθ is θ?

Answer 3

the integral of the cos is the sin

in this case you have
\[ \int \cos(u) dθ\]
where u = 2θ
du = 2 dθ
so we need to put in a factor of 2. compensate with a ½ out front
\[ \frac{1}{2}\int \cos(u) 2 dθ \\ \frac{1}{2}\int \cos(u) du\\
\frac{1}{2} \sin(u) \\
\frac{1}{2} \sin(2\theta)
\]

Answer 4

Oh, my bad. The integral is no longer there when when its at step two.

Answer 5

No wonder I couldn't figure it out! lol.

Answer 6

No, now that makes complete sense, and I could have done that on my own. But thank you for making me realize I wrote down the steps wrong :)

Answer 7

In your post, you show the an integral, but I think that it should not be there.
In other words
\[ \frac{25}{2}\int (1-cos(2\theta)) \; d\theta = \frac{25}{2} (\theta- \frac{1}{2}\sin(2\theta)) \]

Answer 8

You are correct, it should not be there