# UN---Integratable..

$\int\frac{1}{sin^3x+cos^3x}$ Note by Anik Kumar Bhushan
5 years, 3 months ago

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There is a much shorter way to this integral. We have :

$I=\int { \frac { dx }{ { sin }^{ 3 }x+{ cos }^{ 3 }x } } =\int { \frac { dx }{ (sinx+cosx)(1-sinxcosx) } }$

$\Rightarrow I=\int { \frac { (sinx+cosx)dx }{ (1+sin2x)(1-\frac { sin2x }{ 2 } ) } }$

Note : I have used ${(sinx+cosx)}^{2}=1+sin2x$ and $2sinxcosx=sin2x$

Put $sinx-cosx=t,(sinx+cosx)dx=dt$

Also ${t}^{2}=1-sin2x$ hence $sin(2x)=1-{t}^{2}$

$\Rightarrow I=2\int { \frac { dt }{ (2-{ t }^{ 2 })(1+{ t }^{ 2 }) } } =\frac { 2 }{ 3 } (\int { \frac { dt }{ 1+{ t }^{ 2 } } } +\int { \frac { dt }{ 2-{ t }^{ 2 } } } )$

$\Longrightarrow I=\frac { 2 }{ 3 } (\int { \frac { dt }{ 1+{ t }^{ 2 } } } +\frac { 1 }{ 2\sqrt { 2 } } (\int { \frac { dt }{ \sqrt { 2 } -t } } +\int { \frac { dt }{ \sqrt { 2 } +t } } ))$

Finally integrating we have :

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

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

Putting back $t=sinx-cosx$ we have :

$I=\frac { 2 }{ 3 } { tan }^{ -1 }(sinx-cosx)+\frac { 1 }{ 3\sqrt { 2 } } ln(\frac { \sqrt { 2 } +sinx-cosx }{ \sqrt { 2 } -sinx+cosx } )$

- 5 years, 2 months ago

I think your substitution is the combination of my three substitutions...

- 5 years, 2 months ago

Yes,this integral is not so lengthy as it seems to be.

- 5 years, 2 months ago

Now try $\int \frac{1}{(1+x^2)(1+tanx)}dx$

- 5 years, 2 months ago

I am a simple 12th grader who can only do some very basic integrations. I have no idea whatsoever how to do this integral.

- 5 years, 2 months ago

Hehe well if you're a 12th grader then congratz you're very smart if you can do the sin cos integral. Good for you! Oh, and I was joking about that one. It's from HERE

- 5 years, 2 months ago

YES john i think this integral recommends the acknowledgment of some non elementary integration techniques, such as Residue theorem or double integration... which the 12th and 11th graders are not familiar with, unless they try to learn it.

- 5 years, 2 months ago

Ooh actually forget the other one. Try THIS one.

Can you overpower the great W|A?

- 5 years, 2 months ago

HOHO OMG what kinda link was THAT!

Well considering noone on Brilliant posted a solution to this (and I know Brilliant has a ton of people who can do a bunch of crazy stuff), it must be a really hard integral. But, it also must be an integrable integral, considering this function's GRAPH. Looks as bad as a tangent... no complex values or breaks in functions or any sort of weird discontinuities... So yeah, it must be doable.

And while we're at it, can you solve $\int \frac{1}{\sin^3{x}+\cos^3{x}+\tan^3{x}}dx$? Before you try, though, I advise you too look at the answer... but not be too intimidated by it because the answer to your integral came out to be hideous also (but I've got to admit nowhere near as hideous as the one with the tangent).

- 5 years, 2 months ago

heheh woow this one is freaky!! I don't even want to give it a try!! do you know what is the w-shaped letter in the solution???!!

- 5 years, 2 months ago

lol I DUNNO MAN I think it's omega because when I click copyable plaintext it gives a lot of omega^stuff and the only thing that needs a special name there is that $\omega$ (plus it rendered properly under \omega) - but what it means - I got NO clue. @Michael Mendrin ?

- 5 years, 2 months ago

which one is freakier - one with the tangent or the one that W|A can't solve? Even though that one only has the x on top... which somehow makes it harder than adding that extra $\tan^3{x}$ on the bottom.

- 5 years, 2 months ago

from where you come to know about these things?

- 4 years, 9 months ago

@U Z about what exactly? In general I learn from practicing on brilliant and other math websites, also reading articles about analysis, especially those on wikipedia :)

- 4 years, 9 months ago

I read you provided a link to Residue theorem , I was asking these things , I googled Integration but only basics founded which are available in our textbooks.

- 4 years, 9 months ago

@U Z Yes but if you read some more advanced books, such as complex analysis books, you will find new topics: Residue theorem and contour integration are some topics.

- 4 years, 9 months ago

Such as , for 2 months I can't read anything new , but can you suggest me for higher calculus which just contains very basics explanations and only problems.

- 4 years, 9 months ago

@U Z Actually I don't read a textbook, most basic explanations I get are on wikipedia and wolfram . All you have to do is just read and follow a topic that attracts you!

If you want full textbooks, you may try to search on freebookcentre.net for some higher calculus. I will suggest you some books later, when I have time :)

- 4 years, 9 months ago

thnx

- 5 years ago

Our integral can be written as :

$\displaystyle I=\int \mathrm{\frac{1}{(\sin x+\cos x)(1-\sin x\cos x)}}\mathrm{d}x$

Let $x=\frac{u}{2} , dx=\frac{1}{2} du$

$\displaystyle =>I=\frac{1}{2}\int \mathrm{\frac{1}{(\sqrt{\frac{1-\cos u}{2}}+\sqrt{\frac{1+\cos u}{2}})(1-\sqrt{\frac{1-\cos u}{2}}\sqrt{\frac{1+\cos u}{2}})}}\mathrm{d}u$

Rearranging and multiplying the numerator and the denominator by $\displaystyle \sqrt{\frac{1-\cos u}{2}}-\sqrt{\frac{1+\cos u}{2}}$ , we will end up with :

$\displaystyle I=\frac{1}{\sqrt{2}}\int \mathrm{\frac{\sqrt{1+\cos u}-\sqrt{1-\cos u}}{(\cos u)(2-\sin u)}}\mathrm{d}u$

But we have $\displaystyle \sqrt{1+\cos u}-\sqrt{1-\cos u}=\sqrt{(\sqrt{1+\cos u}-\sqrt{1-\cos u})^2}$

$\displaystyle=\sqrt{1+\cos u+1-\cos u -2\sqrt{(1-\cos u)(1+\cos u)}}=\sqrt{2-2\sin u}$

Hence, $\displaystyle I=\int \mathrm{\frac{\sqrt{1-\sin u}}{(\cos u)(2-\sin u)}}\mathrm{d}u$

$\displaystyle =\int \mathrm{\frac{\sqrt{1-\sin u}}{(\cos^2 u)(2-\sin u)}\cos u}\mathrm{d}u$

$\displaystyle = \int \mathrm{\frac{\sqrt{1-\sin u}}{(1-\sin u)(1+\sin u)(2-\sin u)}\cos u}\mathrm{d}u$

Let $\displaystyle y=\sin u , dy=\cos u du$

Therefore, $\displaystyle I=\int \mathrm{\frac{\sqrt{1-y}}{(1-y)(1+y)(2-y)}}\mathrm{d}y$

Now we use method of Partial fractions to get $\displaystyle \frac{1}{(1-y)(1+y)(2-y)}=\frac{1}{2} \frac{1}{1-y}+\frac{1}{6} \frac{1}{1+y} -\frac{1}{3} \frac{1}{2-y}$

and hence $\displaystyle I=\frac{1}{2} \int \mathrm{\frac{\sqrt{1-y}}{1-y}}\mathrm{d}y + \frac{1}{6} \int \mathrm{\frac{\sqrt{1-y}}{1+y}}\mathrm{d}y - \frac{1}{3} \int \mathrm{\frac{\sqrt{1-y}}{2-y}}\mathrm{d}y$

Now, we solve each integral alone:

$\displaystyle A= \int \mathrm{\frac{\sqrt{1-y}}{1-y}}\mathrm{d}y =\int \mathrm{(1-y)^{-\frac{1}{2}}}\mathrm{d}y = -2\sqrt{1-y}$

$\displaystyle B = \int \mathrm{\frac{\sqrt{1-y}}{1+y}}\mathrm{d}y$

Let $\displaystyle t=\sqrt{1-y} , y=1-t^2, dy=-2tdt$

$\displaystyle => B =2\int \mathrm{\frac{t^2}{t^2-2}}\mathrm{d}t = 2\int \mathrm{\frac{t^2-2}{t^2-2}+\frac{2}{t^2-2}}\mathrm{d}t= 2t+4\int \mathrm{\frac{1}{t^2-2}}\mathrm{d}t$

Using method of partial fractions, we get $\displaystyle \frac{1}{t^2-2}=\frac{1}{2\sqrt{2}} (\frac{1}{t-\sqrt{2}}-\frac{1}{t+\sqrt{2}})$ and hence:

$\displaystyle B= 2t+\sqrt{2}\int \mathrm{\frac{1}{t-\sqrt{2}}-\frac{1}{t+\sqrt{2}}}\mathrm{d}t=2t+\sqrt{2}(\ln(t-\sqrt{2})-\ln(t+\sqrt{2}))$

$\displaystyle = 2\sqrt{1-y}+\sqrt{2}(\ln(\sqrt{1-y}-\sqrt{2})-\ln(\sqrt{1-y}+\sqrt{2}))$

$\displaystyle C=\int \mathrm{\frac{\sqrt{1-y}}{2-y}}\mathrm{d}y$

Let $\displaystyle t=\sqrt{1-y}, y=1-t^2, dy=-2tdt$

$\displaystyle => C=-2\int \mathrm{\frac{t^2}{t^2+1}}\mathrm{d}t=-2\int \mathrm{\frac{t^2+1}{t^2+1}-\frac{1}{t^2+1}}\mathrm{d}t=-2t+2\arctan t$

$\displaystyle = -2\sqrt{1-y}+2\arctan\sqrt{1-y}$

Therefore,

$\displaystyle I=\frac{1}{2}A+\frac{1}{6}B-\frac{1}{3}C$ $\displaystyle = \frac{1}{2}(-2\sqrt{1-y})+\frac{1}{6}(2\sqrt{1-y}+\sqrt{2}(\ln(\sqrt{1-y}-\sqrt{2})-\ln(\sqrt{1-y}+\sqrt{2})))-\frac{1}{3}( -2\sqrt{1-y}+2\arctan\sqrt{1-y})$

Now, all what we have to do is substitute back $\displaystyle y=\sin u$ and $x=\frac{u}{2} <=> y=\sin 2x$ :

$=> \sqrt{1-y} = \sqrt{1-\sin 2x} = \sqrt{\cos^2x+\sin^2x-2\sin x \cos x}= \cos x-\sin x$

And by simplifying our answer , we end up with:

$\displaystyle \boxed{\int \mathrm{\frac{1}{(\sin^3 x+\cos^3 x)}}\mathrm{d}x= \frac{\sqrt{2}}{6}\ln(\frac{\cos x-\sin x -\sqrt{2}}{\cos x-\sin x +\sqrt{2}}) -\frac{2}{3}\arctan (\cos x-\sin x) +C}$

- 5 years, 2 months ago

Another way we can write $\sin x+\cos x=\sqrt{2}\sin\left(x+\frac\pi4\right)$ and $1-\sin x\cos x=1-\frac12\sin2x$, then make substitution $t=x+\frac\pi4$. The integral turns out to be $\int\frac{\sqrt{2}}{\left(3-2\sin^2t\right)\sin t}dt.$ Now we can use Weierstrass substitution.

- 5 years, 2 months ago

yes this also works, its great to have more than one way to approach such killer integral !

thanks for sharing your solution :)

- 5 years, 2 months ago

Really you are BRILLIANT.

- 5 years, 2 months ago

Woah. That's a great solution. No complicated transformations, just good old Calc BC stuff.

You should try THIS integral. Solve it and we'll nominate you as the Conqueror of the Integrals!

- 5 years, 2 months ago

Thanks I will try :)

- 5 years, 2 months ago

Oh btw a little typo on on 9th line from below: dt, not dy.

Took me an hour to copy this solution... And how the heck did you come up with this??

$Brilliant.$

- 5 years, 2 months ago

Well.. I tried to come up with a form that fits the method of partial fractions, it seemed that it is the key to this complicated integral!

Btw, the problem you told me about seems to be more tricky...

- 5 years, 2 months ago

hehe don't be upset. you aren't the only person struggling...

109 and counting

- 5 years, 2 months ago

Thanks Hasan I really could'nt solve it. Actually it is an IMO problem and I could'nt come up with a solution......

- 5 years, 2 months ago

welcome :)

- 5 years, 2 months ago

That's easy, it's $x$. Oh, that's cubes, not squares. I need new glasses.

- 5 years, 2 months ago

Ha..haha.

- 5 years, 2 months ago

Thanx to all.....really that's what I call a brilliant site.

- 5 years, 2 months ago

Can you help me solve it

- 5 years, 3 months ago

@Calvin Lin ,Sir can you help me in my post.

- 5 years, 2 months ago

What have you tried? Do you know how to integrate $\frac{ 1}{ \sin x + \cos x }$?

Staff - 5 years, 2 months ago

yes I can solve yours because its raised to power 1, but mine ^3

- 5 years, 2 months ago

Sir Calvin I can't solve.....so can you help me out@Calvin Lin

- 5 years, 2 months ago

Seems like Weierstrass substitution should give you a (potentially nasty) rational function, and those are all integrable by partial fractions.

- 5 years, 2 months ago

Well, your problem is impossible to integrate. But, assuming you meant to integrate with respect to $x$, then

$\int \frac { 1 }{ \sin ^{ 3}{x } +\cos^{3}{x}} { dx }= -\frac { 2}{3 } (\tan ^{ -1 }{ (cos{x}-sin{x}) } +(1+i)(-1)^{3/4}\tanh ^{ -1 }{ (\frac { \tan ({ \frac { x }{ 2 })} -1 }{ \sqrt { 2 } } )} )+C$

- 5 years, 2 months ago

Hasan showed that nothing is impossible.....

- 5 years, 2 months ago

Not what I meant. You forgot to add dx at the end of the integral. The problem makes absolutely zero sense conceptually if you do not add the respect of integration.

But yes Hasan's solution was really elegant :)

- 5 years, 2 months ago