Hard integration question (1 Viewer)

[bradc1988]

How would you integrate cosx.sin^2x? I've tried all the expansions I know but I end up with a cos^3x and other things that do not work.

 

[香港!]

How would you integrate cosx.sin^2x? I've tried all the expansions I know but I end up with a cos^3x and other things that do not work.

let u=sinx then do it..
answer is sin³x\3+C

 

[100percent]

How would you integrate cosx.sin^2x? I've tried all the expansions I know but I end up with a cos^3x and other things that do not work.

i'm assuming you mean cosx*sin²x?
if so, you simply let u=sinx
du=cosx dx
u² du
integrate you get
u³/3
=[sin³x]/3 + c

 

[bradc1988]

let u=sinx then do it..
answer is sin³x\3+C

Hm comes out easy that way. But I got this in my 3 unit trial and I thought if it involves substitution in 3 unit we are told it's substitution.

 

[香港!]

Hm comes out easy that way. But I got this in my 3 unit trial and I thought if it involves substitution in 3 unit we are told it's substitution.

well u dont' need substitution really..
int. cosxsin²x dx
observe that d\dx sinx=cosx
then answer is sin³x\3+C

 

[100percent]

?? you do use subing

 

[香港!]

^
int. cosxsin^2 x dx
=int. sin^2 x d(sinx)
=sin^3 x \ 3+C

 

[justchillin]

Lol...people chain rule like d/dx: (2x^2-1)^2 = 2 (2x^2-1).4x

 

[100percent]

i like to prefer that as "function of a function" chain rule is really when you have multi variables like da/dy= da/db * db/dc * dc/dd * dd/dy

 

[Abtari]

i think this hard integration question is answered sufficiently

 

[Trev]

bradc1988; the same question was in today's exam - did you get it out?

 

[MarsBarz]

Haha I got it out thanks to this thread. Really, what are the odds that a question posted here 2 days before the exam actually appears in the exam hehehe!!

Well thanks a lot anyway !

 

[icycloud]

Alternative method (yes it's much longer and drawn out, but it's always interesting to look at alternative methods ):

I = ∫cos(x) sin²(x) dx
= ∫cos(x)[1-cos²(x)] dx
= ∫cos(x)-cos³(x) dx
= -∫cos³(x)-cosx dx

Now, cos(2x)=2cos²(x)-1
cos(2x)cos(x)=2cos³(x)-cos(x)

cos(2x+x)+cos(2x-x)
=cos(3x)+cos(x)
=2cos(2x)cos(x)

Thus, 2cos³(x)-cos(x)=cos(2x)cos(x)
= [cos(3x) + cos(x)]/2

Thus, I = -1/2 ∫ 2cos³x - cos(x) - cos(x) dx
= -1/2 ∫cos(2x)cos(x) - cos(x) dx
= -1/4 ∫ cos(3x) - cos(x) dx
= -1/4 [1/3 sin(3x) - sin(x)]
= -1/12 sin(3x) + sin(x)/4 + C #

(which is equivalent to sin³x/3 + C)