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Help me integrate sin^2(2x)cosx

Basically the above question please. I cant figure it out, i dont knwo if its a simple integration or much more complicated.

Thanks :smile:

(edited 13 years ago)

Reply 1

Farhan.Hanif93

Original post by Ramin Gorji

Basically the above question please. I cant figure it out, i dont knwo if its a simple integration or much more complicated.

Thanks :smile:

It's simple. Use a substitution of

u=\sin (2x)

if you can't do it by inspection.

Reply 2

Ramin Gorji

Original post by Farhan.Hanif93

It's simple. Use a substitution of

u=\sin (2x)

if you can't do it by inspection.

oops sorry i meant sin^2(2x)cosx

Reply 3

fidan134

if you differntiate sin^3(2x) you get y=3*2sin^2(2x)*cos(2x)=6sin^2(2x)cos(2x)

so when you integrate sin^2(2x)cos(2x), you get (sin^3(2x)/6

Reply 4

Farhan.Hanif93

Original post by Ramin Gorji

oops sorry i meant sin^2(2x)cosx

That complicates things. There are two ways that jump out at me immediately (although it is late, so forgive me if there is a quicker way):

1) Note that

\sin ^2(2x) = \dfrac{1}{2}[1-\cos (4x)]

so our integrand is now

\dfrac{1}{2}\cos x - \dfrac{1}{2}\cos 4x \cos x

. Then note that

\cos 4x \cos x = \frac{1}{2}[\cos (4x + x) +\cos (4x - x)]

.

2) Note that

\sin ^2(2x) \cos x = 4(1-\cos ^2x) \cos ^3x

and expand everything out. The only "difficulty" then is to integrate cos^5x, but that just involves noting that

cos ^5x = (1-\sin ^2x)^2 \cos x

and using a substitution of

u=\sin x

Reply 5

Farhan.Hanif93

Original post by fidan134

if you meant sin^2(2x )cos(x), then use your double angle formula
so you get: 2sin(x)cos^2(x).

That's not right.

Reply 6

Ramin Gorji

Original post by Farhan.Hanif93

That complicates things. There are two ways that jump out at me immediately (although it is late, so forgive me if there is a quicker way):

1) Note that

\sin ^2(2x) = \dfrac{1}{2}[1-\cos (4x)]

so our integrand is now

\dfrac{1}{2}\cos x - \dfrac{1}{2}\cos 4x \cos x

. Then note that

\cos 4x \cos x = \frac{1}{2}[\cos (4x + x) +\cos (4x - x)]

.

2) Note that

\sin ^2(2x) \cos x = 4(1-\cos ^2x) \cos ^3x

and expand everything out. The only "difficulty" then is to integrate cos^5x, but that just involves noting that

cos ^5x = (1-\sin ^2x)^2 \cos x

and using a substitution of

u=\sin x

Thanks for the help :biggrin:

, i preferred your first method seemed a little quicker to do :P

Reply 7

skipp

Just realised however much I like my method it was wrong, damn 3am :emo:

(edited 13 years ago)

Reply 8

Farhan.Hanif93

Original post by skipp

This is a nice little technique for integrating any equation of the form sin^n(x)cos(x), it can easily be adapted for equations of the form cos^n(x)sin(x):

Using sin^2(x) as our example we can integrate this in the following way:

Unparseable latex formula:

\int sin^2\(x\ cos\(x \ dx

(A)

[br]u=sin(x)

(B)

differentiate:

[br]du/dx = cos(x)[br]

rearrange (bad habit here, but it works for this):

[br]du = cos(x) dx

[br][br]\int u^2\ du[br][br]

which you should be able to do easily :smile:

as you can see this method can easily be adapted for any power of sin (x), or for multiple powers of cos(x)with a single sin(x) so it's nice to get a good grip on it, you should also be able to see how to apply this to your own question

His integral isn't of that form, is it? So it doesn't really apply here. :p:

Nonetheless, thumbs up for a good explanation. :wink:

Reply 9

skipp

Original post by Farhan.Hanif93

His integral isn't of that form, is it? So it doesn't really apply here. :p:

Nonetheless, thumbs up for a good explanation. :wink:

Quickly went back and got rid of my post :colondollar:

but not quickly enough it seems

Reply 10

Farhan.Hanif93

Original post by skipp

Quickly went back and got rid of my post :colondollar:

but not quickly enough it seems

Haha, I suppose it's a bit too late for maths anyway; my bed is definitely calling right now. :yawn:

Reply 11

skipp

Original post by Farhan.Hanif93

Haha, I suppose it's a bit too late for maths anyway; my bed is definitely calling right now. :yawn:

Mine too

night farhan :wavey:

hope all this helps OP, despite my tangent

Reply 12

Farhan.Hanif93

Original post by skipp

Mine too

night farhan :wavey:

hope all this helps OP, despite my tangent

G'night skipp. :smile:

Reply 13

slevydc

Original post by Farhan.Hanif93

That complicates things. There are two ways that jump out at me immediately (although it is late, so forgive me if there is a quicker way):

1) Note that

\sin ^2(2x) = \dfrac{1}{2}[1-\cos (4x)]

so our integrand is now

\dfrac{1}{2}\cos x - \dfrac{1}{2}\cos 4x \cos x

. Then note that

\cos 4x \cos x = \frac{1}{2}[\cos (4x + x) +\cos (4x - x)]

.

2) Note that

\sin ^2(2x) \cos x = 4(1-\cos ^2x) \cos ^3x

and expand everything out. The only "difficulty" then is to integrate cos^5x, but that just involves noting that

cos ^5x = (1-\sin ^2x)^2 \cos x

and using a substitution of

u=\sin x

Can you keep going with method #2? I'd love to see it...

Reply 14

old_engineer

Original post by slevydc

Can you keep going with method #2? I'd love to see it...

Do you seriously expect this user to resurrect their working after a gap of nine years?

Reply 15

DFranklin

Original post by slevydc

Can you keep going with method #2? I'd love to see it...

Original post by old_engineer

Do you seriously expect this user to resurrect their working after a gap of nine years?

Here's something "in the spirit of" method 2 but slightly shorter:

Motivation: it's "easy" to find

\int \sin^n x \cos x \,dx

(*) by the substitution u = sin x (or by recognition).

With this in mind, we go:

\int \sin^2 (2x) \cos x \,dx = 4 \int \sin^2 x \cos^2 x \cos x \,dx

= 4 \int \sin^2(1-\sin^2 x) \cos x \,dx = 4 \int \sin^2 x \cos x - \sin^4 x \cos x\, dx

Now both terms are of the form (*), so you can either write down the integral if familiar enough with the pattern, or use the substitution u = sin x to finish.

If you really want to complete method 2, it will be a very similar argument.