how do you integrate cos2x cosx?

does anybody know?

i know you can integrate (cosx)^2 cosx by substitution, but what about cos2xcosx

cos(2x+x)=cos2x cosx - sin2x sinx

cos(2x-x)=cos2x cos x + sin2x sin x

so you have two equations cos3x and cos x. Add them together you get

2cos2xcosx = cos3x + cosx

divide both sides by two

1/2 (cos3x + cosx) then integrate

does anybody know?

i know you can integrate (cosx)^2 cosx by substitution, but what about cos2xcosx

to get 1/3sin3x + sinx?

what are the two equations exactly? :O

and thanks

By parts?

EDIT: Previous method is prettier. Either way is good.

The two equations are the trigonemtric identities from C3.

No the if you integrate 1/2 (cos3x + cosx) you get 1/6 sin3x + 1/2sinx +c.
The 1/2 is outside the brackets.

and can anybody please confirm that d(2^x)/dx = xln2?

I don't think you can do it by parts can u?

Alternatively to the method above, note that $2^x = e^{\ln x^2} = e^{x\ln 2}$ using the following laws of logs:

$a^{\log _aN} = N$

$\log _a b^c = c\log _a b$

Now you can differentiate that like you would normally differentiate e functions:
$\frac{d}{dx} \left[e^{x\ln 2}\right] = \ln 2 \times e^{x\ln2} = \ln 2 \times 2^x = 2^x\ln 2$ as required. So no, it's not xln2.

u = cosx, du/dx = -sinx
dv/dx = cos2x, v = 1/2sin2x

I = 1/2sinxcosx + integral of 1/2sin2xsinx
I = 1/4sin2x + 1/2(Integral 2)

Integral 2:
u = sinx, v = cosx
dv/dx = sin2x, v = -1/2cos2x

I2 = -1/2cos2xsinx + 1/2I

Therefore

I = 1/4sin2x -1/4cos2xsinx + 1/4I
3/4I = 1/4sin2x - 1/4cos2xsinx

so I = 1/3(sin2x - cos2xsinx).

does anybody know?

i know you can integrate (cosx)^2 cosx by substitution, but what about cos2xcosx

u = cosx, du/dx = -sinx
dv/dx = cos2x, v = 1/2sin2x

I = 1/2sinxcosx + integral of 1/2sin2xsinx
I = 1/4sin2x + 1/2(Integral 2)

Integral 2:
u = sinx, v = cosx
dv/dx = sin2x, v = -1/2cos2x

I2 = -1/2cos2xsinx + 1/2I

Therefore

I = 1/4sin2x -1/4cos2xsinx + 1/4I
3/4I = 1/4sin2x - 1/4cos2xsinx

so I = 1/3(sin2x - cos2xsinx).

thanks again..

but how would YOU integrate cos2x cosx?

in the first part, isn't v*du/dx negative?

Well if you note that $\cos 2x = 1 - 2\sin ^2x$ and rewrite the integral using this, you get:

$\displaystyle\int (1-2\sin ^2x)\cos xdx$

Now let $u=\sin x \implies \frac{du}{dx} = \cos x \implies du = \cos xdx$

Then by using that sub you can replace the cosxdx with du because they are equivalent according to the above so you get:
$\displaystyle\int (1-2u^2) du$ and that's easy to integrate.

I can't believe all the mentions of parts and factor formulae or multiplying cosx into it. This is by far the simplest.

Well if you note that $\cos 2x = 1 - 2\sin ^2x$ and rewrite the integral using this, you get:

$\displaystyle\int (1-2\sin ^2x)\cos xdx$

Now let $u=\sin x \implies \frac{du}{dx} = \cos x \implies du = \cos xdx$

Then by using that sub you can replace the cosxdx with du because they are equivalent according to the above so you get:
$\displaystyle\int (1-2u^2) du$ and that's easy to integrate.

I can't believe all the mentions of parts and factor formulae or multiplying cosx into it. This is by far the simplest.

hey i used the same identity as yours, but think i found an even simpler method:

cos2xcosx = (1-2sin^2x)cosx

= cosx-2sin^2cosx

and then integrating, can use reverse chain rule for 2sin^2xcosx

= sinx +2/3(sin^3(x))

am i right?

x