Quick Question (1 Viewer)

[allGenreGamer]

A quick maths question:

Integrate cos(^3)x

How? I know that u start off by spliting it into cosx.cos(^2)x but then I'm lost. The answer is (1/4)((1/3)sin3x + 3sinx + C). Thanks.

 

[Harimau]

( I didn't actually do a calculation to check, i am lazy..)

both answers could be valid, depending on the method.

Slide Rule used the "reverse chain rule" when intergrating it.

Meanwhile, the solution that Gamer gave must have been derived from

Sin(a)*Cos(B) = 1/2 [ Sin(A +B) + Sin (A-B)]

Just use this formula and intergrate normally, you should get the answer that was given.

 

[kpq_sniper017]

To integrate cos³x:

You have to first expand cos3x.
After expanding, cos3x = 4cos³x - 3cosx

.'. 4cos³x = cos3x + 3cosx
.'. cos³x = ¼(cos3x + 3cosx)

and then you integrate from there.

 

[kpq_sniper017]

i'm not too sure if the other ways work.
the only downside with using my method is having to expand cos3x. it's 6 lines of working, so it's not too bad.

 

[grimreaper]

Originally posted by pcx_demolition017
i'm not too sure if the other ways work.
the only downside with using my method is having to expand cos3x. it's 6 lines of working, so it's not too bad.

I think its actually a lot quicker to write cos(^3)x as cosxcos(^2)x, then use cos(^2)x = 1 - sin(^2)x. This ends up as cosx - sin(^2)xcosx, which is easy to differentiate. Also, that method can be used relatively easily for any odd powers of cosx, while doing say cos(^5)x or cos(^7)x using your method would be painful.

 

[shkspeare]

how did u come with the conclusion of expanding cos3x? worked ur way back to get to the answer provided there?

in my opinion... i would just use cosx(1-sin^2x) ...

takes 3 lines .. compared to ur... 8-10 ...

 

[:: ck ::]

the other way happens to be a hell lot shorter.. lol

its just a u substitution [fn and its derivative]

 

[kpq_sniper017]

Originally posted by grimreaper
I think its actually a lot quicker to write cos(^3)x as cosxcos(^2)x, then use cos(^2)x = 1 - sin(^2)x. This ends up as cosx - sin(^2)xcosx, which is easy to differentiate. Also, that method can be used relatively easily for any odd powers of cosx, while doing say cos(^5)x or cos(^7)x using your method would be painful.

lol.
btw. how do u integrate sin²xcosx? using integration by substitution?

 

[kpq_sniper017]

Originally posted by :: ryan.cck ::
the other way happens to be a hell lot shorter.. lol

its just a u substitution [fn and its derivative]

ah well.
it's better to have all methods rather than just one
to be honest, i actually didn't think of breaking it down into cosxcos²x ... i guess i just missed the obvious.

 

[kpq_sniper017]

Originally posted by shkspeare
how did u come with the conclusion of expanding cos3x? worked ur way back to get to the answer provided there?

in my opinion... i would just use cosx(1-sin^2x) ...

takes 3 lines .. compared to ur... 8-10 ...

yea, i probably would too in an exam. as u can see from my post above, i didn't think of doing it that way until i saw this thread.

as for finding out to use cos3x - i've done a question like this before...it works for sin³x too.

 

[Heinz]

Originally posted by pcx_demolition017
lol.
btw. how do u integrate sin²xcosx? using integration by substitution?

letting u=sinx and noting that du/dx = cosx, the integral of sin²xcosx simply becomes the integral of u² du which is easily integrated.