Question

Integral trig, went wrong somewhere I think.

Answer 1

\[\int\limits_{0}^{\pi}2\sin^4x dx\]

Answer 2

Use the standard identities.

Answer 3

2\[\int\limits_{0}^{\pi}\frac{ (1-\cos2x)^2 }{2 }\]

Answer 4

oh sweet love these

start with \[2\int\limits \sin^4 xdx\]

Answer 5

\[2\int\limits_{0}^{\pi} \frac{ (1-2\cos2x + \cos^2(2x))dx }{ 4 }\] I think this is still right soo far.

Answer 6

\[1\int\limits_{0}^{\pi} \frac{ (1-2\cos2x + \cos^2(2x))dx }{ 2 }\] Can I simplify to 1/2 like that?

Answer 7

\[\frac{ 1 }{ 2 } \int\limits_{0}^{\pi} (1-2\cos2x + \cos^2(2x))dx\]

Answer 8

Or do I need to do it like this:
\[2\int\limits_{0}^{\pi} (\frac{ 1 }{ 4 } - \frac{ 2\cos2x }{ 4 } + \frac{ \cos^2(2x) }{ 4 })dx\]