Question

Show that
\[\int\limits_{{1 \over6}\pi}^{{1 \over 3}\pi} \sin3xsinxdx={1 \over 8} \sqrt 3\]

Answer 1

is not like that

Answer 2

in term of maclaurin

Answer 3

What do you mean?

Answer 4

looks like it's going to be ugly.

Answer 5

Can you integrate the sin3xsinx?

Answer 6

smplify sin3x first
http://answers.yahoo.com/question/index?qid=20071210104537AA7XFK9

Answer 7

sin(a)sin(b)=(1/2)[cos(a-b)-cos(a+b)]
this should do it

Answer 8

ghass, that's correct! :D

Answer 9

1(cos 2x − cos 4x) ≡ sin 3x sin x.
But I thought you could integrate with the sin3x sinx alone?

Answer 10

1/2*

Answer 11

yep ... i agree .. elegant solution

Answer 12

1/2(cos 2x − cos 4x) ?

Answer 13

yes,put the limits and you'll get the answer

Answer 14

This isn't the integral though... this is just the factor.

Answer 15

integtrate it .. individually

Answer 16

Can you show how?

Answer 17

I=(1/4)sin(2x)-(1/8)sin(4x)+C

Answer 18

Could you show the steps?