Can you take it from here?
And you know how to integrate that?
What I thought was using what I believe to be some sort of the "power rule".
When one does
so, one can apply that here:
Is this thought process, correct, or have I made any mistake?
Thanks in advance.
Yes, that is correct. Let u= sin(x). Then du= cos(x) dx so your integral is
Note that you could also let u= cos(x). Then du= -sin(x)dx so your integral is
Since , . The two solutions are the same with C= C'- 1/2.
Henryt999 was pointing to a completely different but still valid solution. Since sin(2x)= 2sin(x)cos(x), sin(x)cos(x)= (1/2)sin(2x) and your integral becomes . Now let u= 2x so that du= 2 dx and dx= (1/2)du. Your integral is
But, now, so that solution is the same as . Replacing with , that becomes , the same as our second solution with and so the same as our first solution with .
Thanks HallsofIvy for your very clear explanation. I get it you know.