Integrating 1/[(sin^3x)+(cos^3x)] dx: Step-by-Step Solution

[Vyse007]

Homework Statement

I am asked to integrate 1/[(sin$^{3}$x)+(cos$^{3}$x)] dx. Indefinite integration.

Homework Equations

The Attempt at a Solution

I tried pretty much everything I could think of. Wrote the integral in terms of sin3x and cos3x, and tried to simplify. Even tried a few more identities, and representation in complex form. Found this prob in an undergrad book, so I didn't try DUIS and such. After simplifying I get stuck and I have no idea how to proceed further.

Any help would be appreciated.

 

[JHamm]

Try starting with partial fractions

 

[ardie]

ehem i deleted my own message, woop.
for me it seems to work after changing the denominator to u, and then partial fractions, then some hefty integration which can be looked up i believe.

 

[darkxponent]

substitute : tan (x) = t and then try to solve it

 

[Vyse007]

OK I substituted sinx=u, and then got the whole integral in terms of u. Splitting it into partial fractions, I get two terms: 1/$\sqrt{1-u^{2}}$ and another term 1/u$^{3}$+(1-u$^{2}$)$^{3/2}$. Am stumped about the second term. Actually I don't even know what I did is right!

 

[ardie]

let u = (sin^3 (x) + cos^3(x))

 

[Vyse007]

Well if I put the denominator as u, then du=[3sin^2(x)cos(x)-3cos^2(x)sin(x)]dx, right? Now what? Do I take the whole term out and perform integration w.r.t u? That gives some unexpected answers...
Sorry I haven't brushed up my integration in a while, so please bear with me.
Btw just for fun, I entered this integral in WolframAlpha(tried it for the first time), and the result is not pretty, containing complex terms and inverse tangent hyperbolic functions!

 

[ardie]

scrap that. let tan(x) = u, substitute and solve using the residue theorem.
edited the theorem name

 

[Vyse007]

Hmmm...did try that. Putting tan(x)=u, I get integral (1-u$^{2}$)$^{3/2}$/(1+u$^{2}$)(1+u$^{3}$).
I have no idea how to use residue theorem here, considering that the region has not been specified. Should I simply take up a unit circle to be the region? Also, if that were the case then I would end up with complex terms I guess.

 

[ardie]

if you end up with complex terms then you are doing something wrong

 

[Vyse007]

That doesn't answer my question of how to select the region when none is given. If it were arbitrary, wouldn't the answer be dependent on the region?

 

[ardie]

apart from an abitrary multipicative factor, the region over which you integrate will make no difference to the final result

 

[Vyse007]

apart from an abitrary multipicative factor, the region over which you integrate will make no difference to the final result

I see where you are getting at. But unfortunately, I have no idea how to use residue theorem here, especially when you say that the answer won't have complex terms.
Thanks for helping me out though. Much appreciated!

 

[ehild]

Try this way (although it will be still very ugly):

sin³(x)+cos³(x)=(sin(x)+cos(x))(sin²(x)+sin(x)cos(x)+cos²(x))
And use that sin(x)+cos(x) = √2 sin(x+π/4).

ehild

 

[Bohrok]

You could try the substitution u = tan(x/2) where sinx = 2u/(u² + 1) and cosx = (1 - u²)/(u² + 1). Looks like it will turn into some nasty partial fractions though, but it shouldn't have any complex terms in it.

 

[Vyse007]

Hmm...tried that out and did get some pretty ugly fractions. Got stuck at a point, but that's nothing I can't figure out when I get free.
Thanks for all your help guys!

 

[ehild]

sin³(x)+cos³(x)=(sin(x)+cos(x))(sin²(x)+sin(x)cos(x)+cos²(x))
sin(x)+cos(x) = √2 sin(x+π/4).

The denominator becomes √2 sin(x+π/4)(1+0.5 sin(2x))

Introduce the new variable u=x+π/4.

sin(2x)=sin(2u-pi/2)=-cos(2u).
1+0.5 sin(2x)=0.5(2-cos(2u))=0.5(1+2sin²(u))

The new form of the integral is

$$I=\int{\frac{\sqrt{2}}{sin(u)(1+2 sin^2(u))}du}$$

Substitute $$sin(u)=\frac{2 tan(u/2)}{1+tan^2(u/2)}$$,
the integrand becomes

$$\frac{(1+tan^2(u/2))^3}{\sqrt{2}tan(u/2)(1+10tan^2(u/2)+tan^4(u/2))}$$

Let be t a new variable:

t=tan(u/2), du = 2/(1+t²)dt.

$$I=\int{\sqrt{2}\frac{(1+t^2)^2}{t(t^4+10t^2+1)}dt}= \sqrt{2}\int{(\frac{1}{t}-\frac{8t}{t^4+10t^2+1})dt}$$

It is straightforward to proceed further.

Could somebody check it, please?

ehild

 

[ardie]

sin³(x)+cos³(x)=(sin(x)+cos(x))(sin²(x)+sin(x)cos(x)+cos²(x))

ehild

this equality is incorrect unless the sign of sin(x)cos(x) is minus.
at the very end then how would you go about integrating an integrand which has a polynomial of higher degree in the denominator than the numerator?

 

[ehild]

You are right, I made that mistake while doing the derivation second time (first time it was minus).
This will change the integral to

$$I=\int{\frac{\sqrt{2}}{sin(u)(1+2 cos^2(u))}du}$$

unless I am mistaken again.

ehild

 

[tiny-tim]

using ehild's substitution u = x + π/4, sinu = (sinx + cosx)/√2, sin2x = -cos2u, du = dx …

∫ dx/(sin³x + cos³x)

= ∫ dx/((sinx + cosx)³ - 3sinxcosx(sinx + cosx))

= ∫ du/sinu(2√2sin²u + (3/√2)cos2u)

= ∫ du/sinu(3/√2 + (2√2 - 3√2)sin²u)

= ∫ du/√2sinu(3/2 - sin²u)

= ∫ du/√2sinu(√(3/2) - sinu)(√(3/2) + sinu)

then use partial fractions to give three separate integrands

= ∫ du/√2sinu + (1/2√2)∫ du/(√(3/2) - sinu) - (1/2√2)∫ du/(√(3/2) + sinu)

(eugh … that needs checking )

then use tan(u/2) = t

 

[ardie]

i was told not to do peoples homework, but since everyones writing the solutions i might as well ...
all of this algebra and a final integral which will prove even more challenging in the algebra. so no one has actually tried putting in u = tan(x) from the start?
then we have for the integrand:
(sec(x) / (tan^3(x) + 1)) . du
hence:
integrand = ((u^2 - 1)^1/2) / u^2 + 1
hence integrand = ( 1 / (u-i)) + ( 1 / (u+i))
which doesn't even involve the residue theorem but it just looked like it did to me at the start. i mean without partial fractions it is really trivial to evaluae the integral as the sum of the residues, but provided our contour is over the real line our integral remains to be evaluated, so my mistake once again.
i am not saying what the next line is but it really is trivial

 

[ehild]

using ehild's substitution u = x + π/4, sinu = (sinx + cosx)/√2, sin2x = -cos2u, du = dx …

∫ dx/(sin³x + cos³x)

= ∫ dx/((sinx + cosx)³ - 3sinxcosx(sinx + cosx))

(eugh … that needs checking )[/SIZE]

You are right, it is just ugly this way! I meant transforming the denominator as follows: :

$$(sin(x)+cos(x))(cos^2(x)-sin(x)cos(x)+sin^2(x))=\sqrt{2}sin(x+\pi /4)(1-sin(2x)/2))$$

u=x+pi/4, x=u-pi/4, dx=du, 2x=2u-pi/2, sin(2x)=-cos(2u)

that changes the denominator to:

$$(\sqrt{2}/2) sin(u)(2+cos2(u))=(\sqrt{2}/2)(2+cos^2(u)-sin^2(u))=(\sqrt{2}/2) sin(u)(1+2 cos^2(u))$$

and the integral becomes

$$I=\int{\frac{\sqrt{2}}{sin(u)(1+2cos^2(u))}du}$$
$$I=\int{\frac{\sqrt{2}sin(u)}{sin^2(u)(1+2cos^2(u))}du}$$

Substitute cos(u)=t, sin(u)du =-dt

$$I=\int{\frac{-\sqrt{2}}{(1-t^2)(1+2t^2)}dt}$$

$$I=-\sqrt{2} \int{(\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1+2t^2})dt}$$

It is very easy from here. I hope, there are no more errors...

ehild

 

[ehild]

so no one has actually tried putting in u = tan(x) from the start?
then we have for the integrand:
(sec(x) / (tan^3(x) + 1)) . du
hence:
integrand = ((u^2 - 1)^1/2) / u^2 + 1

Is not the integrand (1+u^2)^1/2 / (u^3+1)? I do not see how did you get your formula.


ehild

 

[SammyS]

After looking at a graph of the integrand, it occurred to me to look at the following:

This seemingly simple substitution may help.

Let u = x + π/4

$\displaystyle \frac{1}{\sin^3(u-\pi/4)+\cos^3(u-\pi/4)}$

$\displaystyle =\frac{2\sqrt{2}}{3\sin(u)+\sin(3u)}$

 

[ehild]

SammyS, how do you think to proceed? I do not feel it easier than the original problem. Take a look at my post before my last one, please!

Now I go further:

In terms of t=cos(u)=cos(x+pi/4), the integral is (as I have shown in the post before my last one)

$$I=-\sqrt{2} \int{(\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1+2t^2})dt}$$.

A=B=1/6, C=2/3.

$$I=\frac{-\sqrt{2}}{3}(\frac{1}{2} ln(\frac{1+t}{1-t})+2arctan(\sqrt{2}t))$$

and in terms of u

$$I=\frac{-\sqrt{2}}{3}(\frac{1}{2} ln(cot^2(u/2))+2arctan(\sqrt{2}cos(u)))$$

ehild

 

[ardie]

nope you are right ehild. so even with my method you would end up having to do the same procedure. maybe the algebra is heavier this way but it is intuitive and easy to understand.

 

[SammyS]

SammyS, how do you think to proceed? I do not feel it easier than the original problem. Take a look at my post before my last one, please!

Now I go further:

...

ehild

Yes, you're right.

After working with it a while, what I suggested is not any better to work with.

WolframAlpha is able to give a solution without the imaginary unit for either of the integrands I have in post #26. That's why initially I thought that post might be useful.

 

[ehild]

Yes, you're right.

After working with it a while, what I suggested is not any better to work with.

Can you check my solution, please? :shy:

ehild

 

[ardie]

seems perfectly fine. here you just forgot a sin(u) in the middle but you carried on the right expression on the right so you can just ignore that

$$(\sqrt{2}/2) sin(u)(2+cos2(u))=(\sqrt{2}/2)(2+cos^2(u)-sin^2(u))=(\sqrt{2}/2) sin(u)(1+2 cos^2(u))$$

 

[ehild]

seems perfectly fine. here you just forgot a sin(u) in the middle but you carried on the right expression on the right so you can just ignore that

Thank you! Now I can sleep

I just found that Wolframalpha shows a form without i.

ehild

 

[ardie]

well the thing is the denominator has so many essential singularities that the integral seems almost useless in mathematical applications
did you really have to do it before you could get some sleep? :D

 

[SammyS]

After looking at a graph of the integrand, it occurred to me to look at the following:

This seemingly simple substitution may help.

Let u = x + π/4

$\displaystyle \frac{1}{\sin^3(u-\pi/4)+\cos^3(u-\pi/4)}$

$\displaystyle =\frac{2\sqrt{2}}{3\sin(u)+\sin(3u)}$

ehild,

This is pretty embarrassing! When I was putting together the above quoted post, I looked through the previous posts, thinking that I might have seen a similar substitution, u = x + π/4, particularly in one of your posts.

I guess I was pretty tired and/or had a brain cramp, but I missed the fact that this was the initial substitution you made in your very nice solution. When going over your solution to try to check it, I graphed $\displaystyle \frac{\sqrt{2}}{ sin(u)(1+2 cos^2(u))}\,,$ and saw that it was the same graph as I had obtained for the above expressions.

I would have never posted that final expression -- the one with the sin(3u) .

At any rate, your solution looks good. I have no idea how you got WolframAlpha to get that rather uncomplicated answer. Both my expressions gave a much more complicated answer.

 

[ehild]

well the thing is the denominator has so many essential singularities that the integral seems almost useless in mathematical applications
did you really have to do it before you could get some sleep? :D

Yes, the graph of the integrand looks terrible, and the whole thing has not much sense, but it was challenging. And the next problem could be to find the improper integral between -pi/4 and 3pi/4.
I could not sleep and when I fall asleep at last, I dreamt of the integral and awoke as I recognised (in my dream) that impossible to derive it without pen an paper.

ehild

 

[ehild]

I have no idea how you got WolframAlpha to get that rather uncomplicated answer. Both my expressions gave a much more complicated answer.

It was the first time that I found and used Wolframalpha. I just typed in integral(1/((sin(x))^3+(cos(x))^3))dx, and it gave the expression with i, but scrolling down, there were alternate forms, one of them that relatively simple one, without i.

ehild