Question

Can you help me with this question please? I am integrating 3 - sin2x. Limits, 3pi/2 and pi/2. I have got 3x + 1/2cos2x, but now I need to evaluate the equation. Can anyone show me how I do this please? Thanks

Answer 1

\[\int\limits_{\frac{ \pi }{ 2 }}^{ \frac{ 3\pi }{ 2}} 3 -sin2x dx \]
Use u-substitution for 2x:
u: 2x
du: 2dx
1/2du:dx
Plug the limits into u:
2(3pi/2) = 3pi
2(pi/2) = pi
\[\frac{1}{2}\int\limits_{\pi }^{ 3\pi} 3 -sinu du \]
Now try! :)

Answer 2

you simply substitide x for 3pi/2 first and then substitude x for pi/2.

then subtract the latter from the first.

so (3*(3pi/2)+1/2cos(2*(3pi/2)) - (3*(pi/2)+1/2cos(2*(pi/2))