Question

find the integral

Answer 1

\[\int\limits (\sin(48x))^2\]

Answer 2

\[\frac{1-\cos^2(48x)}{2}\]

Answer 3

\[\frac{1}{2} \int\limits 1-\cos^2(48x)\]
\[\int\limits \frac{1}{2} - \int\limits \frac{\cos^2(48x)}{2}\]

Answer 4

you didn't apply the power-reducing formula correctly in that second post you made...

Answer 5

\[\frac{1}{2} \int\limits dx - \frac{1}{2} \int\limits \cos^2(48x)dx\]

Answer 6

hmm?

Answer 7

the half angle forumla?

Answer 8

yea... that's basically what it is....

Answer 9

\[\frac{1}{2} \int\limits dx - \frac{1}{2} \int\limits \cos2(48x)dx\]

Answer 10

\[\frac{1}{2} \int\limits dx - \frac{1}{2} \int\limits \cos^(96x)dx\]

Answer 11

that's better..:)

Answer 12

:)

Answer 13

u = 96x
1/96du = dx
\[\frac{x}{2} - \frac{1}{192} \int\limits \cos(u)du\]

Answer 14

\[\frac{x}{2} - \frac{1}{192} \sin(u)+c\]
\[\frac{x}{2} - \frac{1}{192}\sin(96x)+c\]

Answer 15

looks good...

Answer 16

:)

Answer 17

correc! :)