@John Muradeli
–
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.
–
Hasan Kassim
·
2 years, 9 months ago

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).
–
John Muradeli
·
2 years, 9 months ago

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@John Muradeli
–
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???!!
–
Hasan Kassim
·
2 years, 9 months ago

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@Hasan Kassim
–
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 ?
–
John Muradeli
·
2 years, 9 months ago

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@Hasan Kassim
–
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.
–
John Muradeli
·
2 years, 9 months ago

@Megh Choksi
–
about what exactly? In general I learn from practicing on brilliant and other math websites, also reading articles about analysis, especially those on wikipedia :)
–
Hasan Kassim
·
2 years, 3 months ago

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@Hasan Kassim
–
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.
–
Megh Choksi
·
2 years, 3 months ago

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@Megh Choksi
–
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.
–
Hasan Kassim
·
2 years, 3 months ago

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@Hasan Kassim
–
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.
–
Megh Choksi
·
2 years, 3 months ago

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@Megh Choksi
–
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 :)
–
Hasan Kassim
·
2 years, 3 months ago

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@John Muradeli
–
I am a simple 12th grader who can only do some very basic integrations.
I have no idea whatsoever how to do this integral.
–
Ronak Agarwal
·
2 years, 9 months ago

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@Ronak Agarwal
–
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
–
John Muradeli
·
2 years, 9 months ago

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@Ronak Agarwal
–
I think your substitution is the combination of my three substitutions...
–
Hasan Kassim
·
2 years, 9 months ago

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 :

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

\(\displaystyle \boxed{\int \mathrm{\frac{1}{(\sin^3 x+\cos^3 x)}}\mathrm{d}x= \frac{\sqrt{2}}{6}\ln(\frac{\cos x-\sin x -\sqrt{2}}{\cos x-\sin x +\sqrt{2}}) -\frac{2}{3}\arctan (\cos x-\sin x) +C}\)
–
Hasan Kassim
·
2 years, 9 months ago

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@Hasan Kassim
–
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.
–
Tunk-Fey Ariawan
·
2 years, 9 months ago

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@Tunk-Fey Ariawan
–
yes this also works, its great to have more than one way to approach such killer integral !

thanks for sharing your solution :)
–
Hasan Kassim
·
2 years, 9 months ago

@John Muradeli
–
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...
–
Hasan Kassim
·
2 years, 9 months ago

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@Hasan Kassim
–
Thanks Hasan I really could'nt solve it. Actually it is an IMO problem and I could'nt come up
with a solution......
–
Anik Kumar Bhushan
·
2 years, 9 months ago

@Anik Kumar Bhushan
–
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 :)
–
John Muradeli
·
2 years, 9 months ago

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Seems like Weierstrass substitution should give you a (potentially nasty) rational function, and those are all integrable by partial fractions.
–
Patrick Corn
·
2 years, 9 months ago

## Comments

Sort by:

TopNewestThere 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 :

\(I=\frac { 2 }{ 3 } { tan }^{ -1 }(sinx-cosx)+\frac { 1 }{ 3\sqrt { 2 } } ln(\frac { \sqrt { 2 } +sinx-cosx }{ \sqrt { 2 } -sinx+cosx } )\) – Ronak Agarwal · 2 years, 9 months ago

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– John Muradeli · 2 years, 9 months ago

Now try \(\int \frac{1}{(1+x^2)(1+tanx)}dx\)Log in to reply

Residue theorem or double integration... which the 12th and 11th graders are not familiar with, unless they try to learn it. – Hasan Kassim · 2 years, 9 months ago

YES john i think this integral recommends the acknowledgment of some non elementary integration techniques, such asLog in to reply

THIS one.

Ooh actually forget the other one. TryCan you overpower the great W|A? – John Muradeli · 2 years, 9 months ago

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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). – John Muradeli · 2 years, 9 months ago

Log in to reply

– Hasan Kassim · 2 years, 9 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???!!Log in to reply

@Michael Mendrin ? – John Muradeli · 2 years, 9 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.Log in to reply

– John Muradeli · 2 years, 9 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.Log in to reply

– Megh Choksi · 2 years, 3 months ago

from where you come to know about these things?Log in to reply

– Hasan Kassim · 2 years, 3 months ago

about what exactly? In general I learn from practicing on brilliant and other math websites, also reading articles about analysis, especially those on wikipedia :)Log in to reply

– Megh Choksi · 2 years, 3 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.Log in to reply

– Hasan Kassim · 2 years, 3 months ago

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.Log in to reply

– Megh Choksi · 2 years, 3 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.Log in to reply

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 :) – Hasan Kassim · 2 years, 3 months ago

Log in to reply

– Ronak Agarwal · 2 years, 9 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.Log in to reply

HERE – John Muradeli · 2 years, 9 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 fromLog in to reply

– Hasan Kassim · 2 years, 9 months ago

I think your substitution is the combination of my three substitutions...Log in to reply

– Ronak Agarwal · 2 years, 9 months ago

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

– Anik Kumar Bhushan · 2 years, 7 months ago

thnxLog in to reply

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:

\(\displaystyle \boxed{\int \mathrm{\frac{1}{(\sin^3 x+\cos^3 x)}}\mathrm{d}x= \frac{\sqrt{2}}{6}\ln(\frac{\cos x-\sin x -\sqrt{2}}{\cos x-\sin x +\sqrt{2}}) -\frac{2}{3}\arctan (\cos x-\sin x) +C}\) – Hasan Kassim · 2 years, 9 months ago

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Weierstrass substitution. – Tunk-Fey Ariawan · 2 years, 9 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 useLog in to reply

thanks for sharing your solution :) – Hasan Kassim · 2 years, 9 months ago

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– Anik Kumar Bhushan · 2 years, 9 months ago

Really you are BRILLIANT.Log in to reply

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

\(Brilliant.\) – John Muradeli · 2 years, 9 months ago

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Btw, the problem you told me about seems to be more tricky... – Hasan Kassim · 2 years, 9 months ago

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– Anik Kumar Bhushan · 2 years, 9 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......Log in to reply

– Hasan Kassim · 2 years, 9 months ago

welcome :)Log in to reply

109 and counting – John Muradeli · 2 years, 9 months ago

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You should try THIS integral. Solve it and we'll nominate you as the Conqueror of the Integrals! – John Muradeli · 2 years, 9 months ago

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– Hasan Kassim · 2 years, 9 months ago

Thanks I will try :)Log in to reply

Thanx to all.....really that's what I call a brilliant site. – Anik Kumar Bhushan · 2 years, 9 months ago

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That's easy, it's \(x\). Oh, that's cubes, not squares. I need new glasses. – Michael Mendrin · 2 years, 9 months ago

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– Anik Kumar Bhushan · 2 years, 9 months ago

Ha..haha.Log in to reply

Can you help me solve it – Anik Kumar Bhushan · 2 years, 9 months ago

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

\(\int \frac { 1 }{ \sin ^{ 3}{x } +\cos^{3}{x}} { dx }= -\frac { 2}{3 } (\tan ^{ -1 }{ (cos{x}-sin{x}) } +(1+i)(-1)^{3/4}\tanh ^{ -1 }{ (\frac { \tan ({ \frac { x }{ 2 })} -1 }{ \sqrt { 2 } } )} )+C \) – John Muradeli · 2 years, 9 months ago

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– Anik Kumar Bhushan · 2 years, 9 months ago

Hasan showed that nothing is impossible.....Log in to reply

But yes Hasan's solution was really elegant :) – John Muradeli · 2 years, 9 months ago

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Seems like Weierstrass substitution should give you a (potentially nasty) rational function, and those are all integrable by partial fractions. – Patrick Corn · 2 years, 9 months ago

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Sir Calvin I can't solve.....so can you help me out@Calvin Lin – Anik Kumar Bhushan · 2 years, 9 months ago

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@Calvin Lin ,Sir can you help me in my post. – Anik Kumar Bhushan · 2 years, 9 months ago

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– Calvin Lin Staff · 2 years, 9 months ago

What have you tried? Do you know how to integrate \( \frac{ 1}{ \sin x + \cos x } \)?Log in to reply

– Anik Kumar Bhushan · 2 years, 9 months ago

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