Integration Help (1 Viewer)

[sasquatch]

Can somebody please help me with this question.

∫sinxcos2xdx. Its probably very easy but i just cant get the answer in the back of the book.. i get (-2/3)cos³x + cosx + x, where the answer in the back is (-1/6)cos3x + (1/2)cosx + c.

Thanks.

 

[onebytwo]

try changing cos2x to 1 - 2sin^2x, then simplify the integrand to get sinx - 2 sin^3x then using the expansion of sin3@, make sin^3@ the subject and place this in your simplified integrand. it should be simple from there.

 

[Trev]

Don't bother using the expansion of sin3x.
∫ sinxcos2x dx
∫ sinx(2cos²x-1) dx
There, and even if you have to use substitution u=cosx.

 

[onebytwo]

Can somebody please help me with this question.

∫sinxcos2xdx. Its probably very easy but i just cant get the answer in the back of the book.. i get (-2/3)cos³x + cosx + x, where the answer in the back is (-1/6)cos3x + (1/2)cosx + c.

Thanks.

did you mean + c

 

[sasquatch]

Can someone actually do the question.. I used subsitution of u = cos²x, and got the answer i put (yeah i meant c) and then i used subs. of cos x as trev suggested...and i get the same answer.

Can somebody else try it..thanks..

 

[Trev]

∫ sinx(2cos²x-1) dx
u=cosx; -du=sinx.dx
Integral becomes:
∫ -(2u²-1) du
= -[(2/3)u³-u] + C
= -(2/3)cos³x + cosx + C

 

[sasquatch]

yeah see thats what i got...thanks...hehe

 

[sasquatch]

Could somebody check my answer to this question (i again got a differnet answer).


I=∫cos2xcos4xdx
=∫cos2x(1-2sin²2x)dx

Let u = sin2x
du/dx = 2cos2x
du/2 = cos2x

.:. I = (1/2)∫(1-2u²)du
= (1/2)(u - 2u³/3) + c
= (1/2)sin2x - (2/6)sin³2x + c

 

[Riviet]

Yep, looks right, should also simplify that 2/6 to 1/3.

 

[sasquatch]

Oh yeah woops...the answers well two so far are wrong..hehe.. Well i dunno if theyre are multiple solutions.. but i dont see how their answer is obtainable...

 

[Riviet]

By the way, is this question from Fitzpatrick? Because that textbook has many mistakes in the answers.

 

[sasquatch]

YES!!! hehehe...

 

[onebytwo]

i think you guys will find that the questions are not at all wrong but just in a different form. see, your answers involve cos^3x, while the answers in the back have cos3x. if you want to check, try expanding cos3x, then from the expansion make cos^3x the subject, and sub it into the answer you got. i tried it for that first question you asked and turns out there exactly the same

 

[onebytwo]

has anyone noticed that in fitzpatricks 33(a), questions 22 and 27 are exactly the same? strange thing is, there presented in the back as having different answers

 

[insert-username]

has anyone noticed that in fitzpatricks 33(a), questions 22 and 27 are exactly the same? strange thing is, there presented in the back as having different answers

I noticed that and had a laugh. 27 has the correct answer for both of them, whereas the answer for 22 is wrong. I remember doing it and thinking "What?" the first time and "Hey, that's familiar" the second time...


I_F