Question

evaluate the integral: sin^2 (k) dx between 0 and 2pi

Answer 1

write it as: \[sin^2(x) = \frac{1-cos2x}{2}\]

Answer 2

sorry its actually : sin^2 (k)x dx between 0 and 2pi

Answer 3

If \(k\) is a constant then treat \(sin^2(k)\) just like you would any other number because it will also be constant. So you will get:
\[\frac{sin^{2}(k)x^2}{2}\] between 0 and \(2\pi\).
Are you able to finish it off?

Answer 4

Also do you see how we get that? It's just like integrating \(3x\), which would be \(\frac{3x^2}{2}\).

Answer 5

wait how did you get x^2

Answer 6

Would you be able to integrate something like \(2x\)?

Answer 7

yes i can

Answer 8

Is \(k\) just a constant in your question btw I'm assuming?

Answer 9

ok

Answer 10

If \(k\) is just a constant then \(sin^2(k)\) will also be a constant, so just treat \(sin^2(k)x\) like you would any other constant multiplying \(x\). Does that make sense?

Answer 11

ok.

Answer 12

Anything I can help with?

Answer 13

nope. i got it. thank you :D

Answer 14

Excellent :)