Integration question (1 Viewer)

[tacogym27101990]

hey can someone help me with this question please?
its probably alot easier than i think it is but yeah..



π/6
∫ sin4xcos2xdx
-π/6



thanks for your help

 

[Slidey]

Use product to sum formula:
sin4xcos2x=(sin6x+sin2x)/2

Can be derived from the addition to product formula:
sin(A+B)=sinAcosB+sinBcosA
Now put in B => -B
sin(A-B)=sinAcosB-sinBcosA

Add the two. Voila.

Oh lol I am a total tool. Remember how sin(a)=-sin(-a)? Well take a look at the terminals of the integral. You'll get an answer of the form F(a)-F(a) because the function to be integrated is odd.

That is: the answer is zero.

 

[webby234]

cos(2x)sin(4x)
= cos2x * 2sin(2x)cos(2x)
=2 cos²(2x) * sin(2x)

Then just use the substitution u = cos2x

EDIT: Slidey's way is probably how it's meant to be done . I think they both work though...

 

[Slidey]

Ah, nice! Yeah I love integration because of the variety of solutions.

 

[tacogym27101990]

thanks guys
makes much more sense
i was trying to use the fact that cos2x = 2cos^2x-1

it got pretty messy

 

[minijumbuk]

Use product to sum formula:
That is: the answer is zero.

Not like I do 4U, nor did I know how to do it, but man my instincts are right xDDD