help help integration =D (1 Viewer)

[chen yi]

Hey I'm stuck on this question. Could somebody help me out? Find the primitive function of (cos x)^3

 

[webby234]

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

u = sin x
du = cosxdx

So Integral of cosx(sinx)^2 dx

= u^2 du

So the integral of (cosx)^3

= sinx - u^3/3
= sinx - ((sinx)^3)/3

Sorry for the poor setting out, but the important part is turning (cosx cubed) into (cosx times cosx squared)

 

[NickP101]

Both good answers, and im not trying to take over this thread, but a question i have been meaning to ask is how do u know when u can use substitution and not. In the first method to integrate cos^3x substitution was used, it gives the correct answer.

What im getting around too is why cant u use substitution when integrating something like cosxsinx? It wont give the correct answer. (u = sinx) (du = cosx).

Leading to my next question, in an exam should u only use substitution when asked to (When the questions is related to trig functions, ive never run into problems with other functions)?

Hope im not asking a stupid question.

Thanks all.

Nick

 

[NickP101]

Thanks Iruka, that was stuffing around with my head.

 

[chen yi]

thanks alot for your help. =D

 

[chen yi]

thanks alot for your help. =D

 

[echelon4]

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

u = sin x
du = cosxdx

So Integral of cosx(sinx)^2 dx

= u^2 du

So the integral of (cosx)^3

= sinx - u^3/3
= sinx - ((sinx)^3)/3

Sorry for the poor setting out, but the important part is turning (cosx cubed) into (cosx times cosx squared)

i'm just a little confused in the working out. Why did you expand (cosx)(1-sin^2x)?

It's a little hard to explain what i'm trying to ask, so i'll just type out my working out.

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

u=sinx, du/dx=cosx...........du=cosxdx

sub in du

so, the integral is (integral of) 1-u^2 du= x-sin^3x/3

isn't it?

 

[rama_v]

so, the integral is (integral of) 1-u^2 du= x-sin^3x/3

isn't it?

The substitution you used was u = sin x
So when you integrate (1-u²)du you get u - u³/3
which is sin x - sin³x / 3

 

[echelon4]

Ahhh, i c now. THanks alot!

 

[MolestLawzy]

sick chen yi

ask me at school