Question

i dont get how they got a final anwer from (3/8)x-(3/32)sin(4x)-(1/8)sin^3(2x)cos(2x)+C ....to (1/64)(24x-8sin(4x)+sin(8x)) +C

Answer 1

What was the original integral?

Answer 2

sin^4(2x)

Answer 3

Give me just a second to run through it.

Answer 4

I got:
\[I=\frac{3}{8}x-\frac{1}{8}\sin(4x)-\frac{1}{64}\sin(8x)+C\]
Which is what wolfram got...

Answer 5

ok... ya i got it .. i missed a step

Answer 6

Errr...that should be +1/64sin(8x)+C...Damn mistypes lol.

Answer 7

ha.. its all good.. thanks though

Answer 8

No problem :P