Integration with constant in function

Question

By right: $$\int\sin\left(\frac\pi2+v\right)\,dv=-\cos\left(\frac\pi2+v\right)+K$$ But when I put this into a computer algebra system, I got back $\sin v$. Why?

Answer 1

Because $-\cos(x + \frac{\pi}{2}) = \sin x$.

Answer 2

Either

(1) use a substitution $t = \frac{\pi}{2} + v$, or

(2) expand $\sin( \frac{\pi}{2} + v) = \sin \frac{\pi}{2} \cos v + \cos \frac{\pi}{2}\sin v=\cos v$.

Answer 3

Using Addition Formula for Sine, $$\sin \left(\frac{\pi}{2} + v\right) = \cos v.$$

Then $$\int \sin \left(\frac{\pi}{2} + v\right) \text{ } dv = \int \cos v \text{ }dv = \sin v + C.$$