\[ \int \frac{ \tan^3 x } { \sin^3 x + \cos^3 x } \, dx \]

If you give up, you can look below.

\[ \int \frac{ \tan^3 x } { \sin^3 x + \cos^3 x } \, dx \]

If you give up, you can look below.

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TopNewest\(\displaystyle \int \frac{\tan^3 x}{\sin^3 x +\cos^3 x} dx= \int \frac{\tan^3 x +1}{\sin^3 x +\cos^3 x} dx - \int \frac{1}{\sin^3 x +\cos^3 x} dx \)

Now working on the first one, take \(\displaystyle \cos^3 x\) as a common factor in the denominator : \(\displaystyle \int \frac{\tan^3 x +1}{\sin^3 x +\cos^3 x} dx = \int \frac{\tan^3 x +1}{\cos^3 x (\tan^3 x +1)} dx = \int \sec^3 x dx \)

Let \(\displaystyle I = \int \sec^3 x dx\) , Integrating by parts with \(\displaystyle u=\sec x , dv= \sec^2 x dx\) :

\(\displaystyle I= \sec x \tan x - \int \sec x \tan^2 x dx= \sec x \tan x - \int \sec x (\sec^2 x -1) dx = \sec x \tan x - \int \sec^3 x dx + \int \sec x dx \)

\(\displaystyle => I= \sec x \tan x - I + \ln(\sec x + \tan x) \)

\(\displaystyle => I= \frac{ \sec x \tan x + \ln(\sec x + \tan x)}{2} \)

Then we have :

\(\displaystyle \int \frac{\tan^3 x}{\sin^3 x +\cos^3 x} dx = \frac{ \sec x \tan x + \ln(\sec x + \tan x)}{2} - \int \frac{1}{\sin^3 x +\cos^3 x} dx\)

You can see the solution to the last integral from This note. Thus:

\(\displaystyle \boxed{\int \frac{\tan^3 x}{\sin^3 x +\cos^3 x} dx = \frac{ \sec x \tan x + \ln(\sec x + \tan x)}{2} - \frac{\sqrt{2}}{6} \ln(\frac{\cos x -\sin x -\sqrt{2}}{\cos x -\sin x +\sqrt{2}}) + \frac{2}{3} \arctan(\cos x -\sin x ) + C} \) – Hasan Kassim · 2 years, 10 months ago

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Hmm... Which one looks more threatening, one with tangent or one with just sin and cos on bottom? Sayin' cause it seems like tan can be used to simplify sin and cos, and plus when I usually see many things I think they can be manipulated in a certain way to be less complex (like this). And when there's less stuff it seems like you have less options to work with and it all looks scary lol... But this is cause I'm experienced, I guess. What about for those who ain't that pro at math? DUnno... Lol.

Well,

King Integral, I've got more stuff coming up for ya. Meanwhile, maintain that streak (we're head to head)! Whoever breaks first is a dweeb.Now get back to being awesome

salsa

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Btw I will try to keep my streak alive although my school began today :) – Hasan Kassim · 2 years, 10 months ago

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the toughest part is not to forget. but yeh, go with that if ya want :D – John Muradeli · 2 years, 10 months ago

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heheh me too I solve level one algeba problem to keep my streak :)) – Hasan Kassim · 2 years, 10 months ago

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– John Muradeli · 2 years, 10 months ago

O_O oooomg no way did you actually do this?? I got no time to check but WHOA and for some reason your answer looks waay shorter than and of WolframAlpha's answers. I'ma check later but great job if you got it rightLog in to reply

– Hasan Kassim · 2 years, 10 months ago

yes I checked it several times,also I will try to find a way to prove that wolfram alpha answers equal this.Log in to reply

right now me gotta speed off somewhere, but I shall address this feat when I return. Think of any cool names for yourself by then ;) I thought of

Intergral MasterorKing Integrallol... – John Muradeli · 2 years, 10 months agoLog in to reply

– Hasan Kassim · 2 years, 10 months ago

Hehehehe :)Log in to reply

– Megh Choksi · 2 years, 10 months ago

how did you type the integral signLog in to reply

between \(

and \) – Hasan Kassim · 2 years, 10 months ago

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8 mins, not bad.

here ya go:

s

And yeah this is doable...

sa

Whoever solves this gets to call itself whatever it wants.

SOLVE ONHere's some extra booze if you need some:

aga

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