Integrating odd power of cosine Help! (1 Viewer)

[dishab]

K i have to integrate cos^3 x dx. The answer i get is Sinx - Sin^3x/3 +C.
But the book says; 1/4( Sin3x/3 + 3 Sinx) +c

 

[tommykins]

maybe you should show us your working?

but the answers would be the same if you expand it out.

cos^3 x = cosx(1-sin^2x)

integral of that is sinx - (sinx)^3/3

 

[dishab]

cos^3 x dx = cos^2x . cosx

=(1-sin^2x).cosx
= let u = sinx
then du = cosx.dx
therefore; (1-u^2) du
=u-u^3/3
= sin- sin^3/3.
wat am i doin wrong?

 

[Iruka]

cos^3 x dx = cos^2x . cosx

=(1-sin^2x).cosx
= let u = sinx
then du = cosx.dx
therefore; (1-u^2) du
=u-u^3/3
= sin- sin^3/3.
wat am i doin wrong?

Nothing, you've used a different trig identity, that's all.

The answer according to BOB comes from using the identity
cos(3A) = 4 cos^3(A) - 3 Cos (A).

If you graph the two answers you will find that they (a most) differ by a constant.

This happens a lot when doing calculus with trig functions.

 

[dishab]

K i am lost.

 

[untouchablecuz]

 

[dishab]

Thank you untouchablecuz and Iruka. i had the right answer :rofl: