@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

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.

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).

@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 ?

@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.

@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 :)

@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.

@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.

@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.

@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 :)

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 \(\displaystyle \frac{1}{(1-y)(1+y)(2-y)}=\frac{1}{2} \frac{1}{1-y}+\frac{1}{6} \frac{1}{1+y} -\frac{1}{3} \frac{1}{2-y}\)

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.

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.

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## Comments

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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 :

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

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

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

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

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

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HERE

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

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.

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THIS one.

Ooh actually forget the other one. TryCan you overpower the great W|A?

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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).

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@Michael Mendrin ?

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

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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 :)

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thnx

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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:

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

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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.

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

thanks for sharing your solution :)

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Really you are BRILLIANT.

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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!

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Thanks I will try :)

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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??

\(Brilliant.\)

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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...

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109 and counting

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

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Ha..haha.

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

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Can you help me solve it

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

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

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

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

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

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

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

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Hasan showed that nothing is impossible.....

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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 :)

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