Trigonometric Integral: Solving \int(cosx)^5sin(x)dx

[icesalmon]

Homework Statement

\int(cosx)^5sin(x)dx, the question asks me to simply find the integral.

Homework Equations

cos^2(x) + sin^2(x) = 1

The Attempt at a Solution

I have an odd power of cosine so I should split that up into [cos^2(x)]^2(sin(x)cos(x)), then I change cos^2(x) to 1-sin^2(x) and let u = sin(x) so I get (1-u^2)^2(u)du factor distribute and then integrate, but this isn't leading me to the correct answer.

 

[Dick]

Just substitute u=cos(x). Where does that take you?

 

[ehild]

Use that sin(x) is the negative derivative of cos(x). Taking y=cos(x), your integral becomes -∫y^5 y' dx.

ehild

 

[icesalmon]

the right answer, but I get a negative out in front. It feels easier to do this the way I have, but is there a reason it's incorrect? What I have done isn't illegal is it? It seems to make this easier to integrate for me, but I get longer answers with different functions. In simplest terms this happens because I'm not doing what you say, but why is your method better suited for this final step? my substitution still cancels the cosine and I have a u to distribute.

 

[HallsofIvy]

As others said, the simplest thing to do is to let u= cos(x) so that du= -sin(x)dx
\int cos^5(x) sin(x)dx= \int u^5(-du)= -\int u^5 du.
= -u^6/6+ c= -cos^6(x)/6+ c

If, instead, you write the integral as
\int (cos^2(x))^2 (cos(x)sin(x)dx)= \int (1- sin^2(x))^2(sin(x))(cos(x)dx)
and let u= sin(x), then du= cos(x)dx
and your integral becomes
\int (1- u^2)^2(u)(du)= \int (u^5- 2u^3 + u)du= u^6/6- u^4/2+ u^2/2+ C
= sin^6(x)/6- sin^4/2+ sin^2/2+ C
= \frac{1}{6}(sin^6(x)- 3sin^4(x)+ 3sin^2(x))+ C
= \frac{1}{6}((1- cos^2(x))^3- 3(1- cos^2(x))^2+ 3(1- cos^2(x)))+ C

Now, "do the algebra". You will find that the cos^4(x) and cos^2(x) terms cancel leaving -cos^6(x)/6 but with a different constant.

 

[icesalmon]

my professor mentioned there being different constants of integration after going back and solving using some initial condition with a similar problem. Which is strange to me because I view these answers as different forms of the same thing. But they aren't the same function at all.

Well I see what dick and ehlid were speaking of now, thanks for going through the trouble of typing all that out. There are guidlines my book points out that I am trying to get used to with these trigonometric integrals.

 

[icesalmon]

well I calculated that they are the same thing, and I have another question.

find \intsec³(\pix)

I've tried two things so far, both of which involve splitting up my secant into sec²\pixsec\pix and sec\pix(tan²\pix + 1) I have used integration by parts on both,

For the first I obtain sec\pixtan\pix/\pi + (1/\pi)²ln|cos\pix| + c

And on the second I obtain 2xsec\pix + (sec\pixtan\pix)/\pi +1/\pi²(ln|cos\pix|) - x² + c

Neither of which are correct.

For the first I set u = sec\pix and dv = sec²\pix so my v = ((tan\pix)/\pi) + c

For the second I, again, set u = sec\pix but I let dv = 1+ tan²\pix so that v = 2x + (1/\pi)(tan\pix) + c

Not sure where I'm going wrong, apologies for lighting up this homework thread so much

 

[talk2glenn]

Forget about the pi*x. Let U=pi*x, dI=pi dx, so you can factor it out by just putting 1/pi outside the integral and ignoring it going forward. Might help you see your own work better.

You are right about integration by parts. Split the integral into factors of sec, and then apply parts. u=secU, du=secUtanU. dv=sec^2U, v=tanU.

When you expand terms, you'll end up with an integral of sec^3 on the right hand side. Notice that this is the same term you are solving for, and move it to the other side of the equation. Divide both sides by two. Everything remaining on the right hand side is now easily integrable.