Integrate this.

$\int \frac{\sin x +\cos x}{\sin^4x + \cos^4x} dx$

Alternatively evaluate the following expression: $\int \frac{1}{2z^4-2z^2+1}dz$

#HelpMe! #MathProblem #Math

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

$\int \frac { \sin x + \cos x } { \sin^4 x + \cos^4 x } dx$

$\sin^4 x + \cos^4 x$ $= (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cdot \cos^2 x$ $= 1 - \frac{1}{2} \sin^2 ( 2 x )$

$\int \frac { \sin x + \cos x } { \sin^4 x + \cos^4 x } dx$ $= 2 \int \frac { \sin x + \cos x } { 2 - \sin^2 ( 2 x) } dx$

Note that $\frac{d}{dx} (\sin x - \cos x) = \sin x + \cos x$

$(\sin x - \cos x)^2 = 1 - \sin (2x)$

$\sin^2 (2x) = (1 - (\sin x - \cos x)^2)^2$

$\int \frac { \sin x + \cos x } { \sin^4 x + \cos^4 x } dx$ $= 2 \int \frac { \sin x + \cos x } {2 - ( 1 - (\sin x - \cos x)^2)^2 } dx$

Let $y = \sin x - \cos x$ , then $(\sin x + \cos x ) dx = dy$

$= 2 \int \frac {1} {2 - ( 1 - y^2)^2 } dy$

$= 2 \int \frac {1} {(\sqrt{2})^2 - ( 1 - y^2)^2 } dy$

$= -2 \int \frac {1} {(y^2 - 1 + \sqrt{2})(y^2 - 1 - \sqrt{2}) } dy$

Split the integrand by Partial Fractions. Evaluate the integral, back substitute and you're done.

Pi Han Goh - 8 years, 1 month ago

after this we can apply forcing integration by parts and then i got the answer....after furthr solving....

A Former Brilliant Member - 8 years, 1 month ago

I did the same thing yesterday.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – go to that post i posted ysterdae....

A Former Brilliant Member - 8 years, 1 month ago

IMPOSSIBLE! No dx anywhere to be found! :D

Kenneth Chan - 8 years, 1 month ago

Now it is there.. Thanks for pointing it out!

Aditya Parson - 8 years, 1 month ago

Hey I got something like this so far... Integral( sint / 1 + (sin2t)^2 dt) .... I am not getting how to proceed from here ...

Saloni Gupta - 8 years, 1 month ago

How did you get sint in the numerator?

Aditya Parson - 8 years, 1 month ago

How did you get this?

Aditya Parson - 8 years, 1 month ago

No solutions. Can the brilliant staff help me out?

Aditya Parson - 8 years, 1 month ago

hey....Aditya m so sorry tried alot but m stuck at the same step over which i was before.......still trying....:(

A Former Brilliant Member - 8 years, 1 month ago

Sure. No problem.

Aditya Parson - 8 years, 1 month ago

A Former Brilliant Member - 8 years, 1 month ago

Hi Aditya, this is a tricky question but it can be solved easily. Just separate the given integral in two different integrals, first whose numerator is sinx plus second integral whose numerator is cosx while the denominator remains the same (sin^4x + cos^4x). now put t=cosx in first integral and u= sinx in second. Both can be done separately to obtaion the answer. Very goooooooood question!!!!!!!11

Kumar Ashutosh - 8 years, 1 month ago

I have tried that method too,but still I have to encounter a weird and complex expression.

Aditya Parson - 8 years, 1 month ago

I got an answer......but dont know how to write it in latex.....what to do???/

A Former Brilliant Member - 8 years, 1 month ago

just write it in fragments, I will understand.

Aditya Parson - 8 years, 1 month ago

i got

-1/2* (1/sinx-cosx) + 1/4 * {ln (sinx-cosx-1/sinx+cosx+1)} +1/(sinx-cosx)(2sin^2cos^2)

is this the answer?

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – No that is not the answer.. But I would like to see your working, as to how you derived the above solution.

Aditya Parson - 8 years, 1 month ago

You can scan the paper where you have worked it out, and post it in the discussions?

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – ok i'll do that but u have to wait as i have to fair it out

will u w8?

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – Yeah I will wait.

Aditya Parson - 8 years, 1 month ago

SIMPLY CLICK HERE

Sayan Chaudhuri - 8 years, 1 month ago

Hi,

How did you get the step before your answer? Did you try differentiating the answer?

gopinath no - 8 years, 1 month ago

NOPE...:().....!!! any integration does not follows by a differentiation ......there was a formula for that very step ....in by text book.....i took resort to that formula.......:)

Sayan Chaudhuri - 8 years, 1 month ago

@Sayan Chaudhuri – Text books can be wrong. What I meant was differentiating your answer and see if it matches with the question.

gopinath no - 8 years, 1 month ago

@Gopinath No – oh...!!...many one do this and even me too...that is the easiest way to check whether it is right or wrong....but Aditya P. pointed out my mistakes.....thanks him....

Sayan Chaudhuri - 8 years, 1 month ago

Wrong.

Aditya Parson - 8 years, 1 month ago

The last steps are WRONG!! Check it yourself.

Aditya Parson - 8 years, 1 month ago

oh...sorry...sorry....i mistook doing hastily...

Sayan Chaudhuri - 8 years, 1 month ago

@Sayan Chaudhuri – Well, the link you provide is too clumsy. I would appreciate it if someone posts the solution with precision.

Aditya Parson - 8 years, 1 month ago

here is the full solution......please let me know if there is any wrong in this solution..........HOPE THIS HELPS..............THE FINAL STEP,YOU CAN SEE,HERE........THANX...

Sayan Chaudhuri - 8 years, 1 month ago

@Sayan Chaudhuri – Also the solution you have proposed is wrong yet. The last step is wrong. If you analyse your solution carefully, you will notice that you have written $sin^2(2x)$ as $[1-(sinx-cosx)^2]$ which is arguably wrong and the correct representation should be : $[1-(sinx-cosx)^2]^2$

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – thanks for pointing out mistakes......sorry for this mistake.....then you have got the answer as ,i see, you have understood how to proceed......once again saying sorry for mistake ....in future i take oath to check my answer throughly ......

Sayan Chaudhuri - 8 years, 1 month ago

@Sayan Chaudhuri – No, I have not yet got the answer.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – I already knew these steps you put up here, but they did not take me anywhere close to the answer.

Aditya Parson - 8 years, 1 month ago

I think you are missing the limits here.

gopinath no - 8 years, 1 month ago

Huh? Indefinite integration!!

Aditya Parson - 8 years, 1 month ago

what limits has to do with this question?

m getting confused...........

A Former Brilliant Member - 8 years, 1 month ago

It has nothing to with limits. It is indefinite integration.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – yeah thats what i thought.................

thanks for clearing my doubt........:)

A Former Brilliant Member - 8 years, 1 month ago

@Aditya Parson – u edited the question few hours back????????

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – Yeah, I did.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – ok

A Former Brilliant Member - 8 years, 1 month ago

If there's limit from 0 to pi/2, we can easily solve it. Indefinite integral has no neat solution, so I asked..

gopinath no - 8 years, 1 month ago

@Gopinath No – I would not have posted the question if it had the limits as you mentioned. That would make this problem easy, but as far as I know that to solve this by indefinite integration is a lengthy process.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – Okay, no problem. My advice would be to ask a computer algebra system when you are stuck. If it does not give a nice answer, then there might be a problem with the question.

gopinath no - 8 years, 1 month ago

@Gopinath No – I integrated it on wolfram alpha and it produced a rather complex answer which I suppose and believe is correct, and my textbook corroborates the answer. But, wolfram alpha fails to display the solution and the steps with much needed clarity. Also, the solution posted by Raja S. is incomplete and also I have pointed out the error.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – Your textbook has an answer? What's the name of the book?

gopinath no - 8 years, 1 month ago

@Gopinath No – RD sharma

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – Okay, I had that book once upon a time. Is it the latest edition ?

gopinath no - 8 years, 1 month ago

@Gopinath No – the same edition you are having can work...[if u r still having R.D. ]

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – Uhmm.. Pardon me?

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – sorry i didnt got what do you mean?

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – Sorry, I mistook your previous comment as directed towards me, so I'd for a moment lost context. Anyways, no solutions for this integral?

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – its ok.....no problem i didnt mind it :)

i didnt tried the problem again actually....first time when i tried it i didnt got the answer

plss dont mind i'll try it again and post the answer by tomorrow if i would be able to get it.....

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – I will wait..

Aditya Parson - 8 years, 1 month ago

@Gopinath No – And yeah the latest edition it is...

Aditya Parson - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$