integral question, wtf is wrong wif my working? (1 Viewer)

[bleakarcher]

integral[sin(4x)sin(2x)] dx from x=0 to x=pi/4
=(1/2)integral[cos(2x)-cos(6x)] from x=0 to x=pi/4
=(1/2)[(1/2)sin(2x)-(1/6)sin(6x)] from x=0 to x=pi/4
=(1/2)[(1/2)-(1/6)]=(1/2)[1/3]=1/6
the answer is 1/3, i can not see wat is wrong wif my working

 

[Riproot]

1/2 * (1/2 - 1/6) = 1/2 * 1/3 = 1/6 not 1/2

 

[Studymates]

Riproot is wrong
the correct way is:

After integration you get:

[1/2sin2x-1/6sin6x] from pi/4 to 0
=[1/2sin(90)-1/6sin(270)]-0
=[1/2(1)-(1/6)(-1)]-o
=1/3
q.e.d

please rep me for my contribution

 

[AAEldar]

Riproot is wrong
the correct way is:

After integration you get:

[1/2sin2x-1/6sin6x] from pi/4 to 0
=[1/2sin(90)-1/6sin(270)]-0
=[1/2(1)-(1/6)(-1)]-o
=1/3
q.e.d

please rep me for my contribution

You shouldn't be putting degrees in when evaluating what's given in radians.

Also don't ask for rep.

 

[bleakarcher]

how is it 1/3 though?
sin(4x)sin(2x)=1/2[2sin(4x)sin(2x)]=1/2[cos(2x)-cos(6x)]

 

[SpiralFlex]

By products to sums, we use:





With the limits.



Doing this slowly,



(It's PLUS)

 

[SpiralFlex]

how is it 1/3 though?
sin(4x)sin(2x)=1/2[2sin(4x)sin(2x)]=1/2[cos(2x)-cos(6x)]

Integrate that and substitute the limits.

 

[Riproot]

Riproot is wrong

M8, I was correcting the last line of his working that was clearly wrong. You retarded? I'm on mah phone and cbf looking at the rest, but I assumed if he fucked that up he would've fucked up the rest too.

 

[bleakarcher]

yeh thanks guyz, sin(3pi/2)=-1, lol