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Integrate this.

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

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

Note by Aditya Parson
4 years, 5 months ago

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

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

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

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 - 4 years, 5 months ago

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

Riya Gupta - 4 years, 5 months ago

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I did the same thing yesterday.

Aditya Parson - 4 years, 5 months ago

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

Riya Gupta - 4 years, 5 months ago

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No solutions. Can the brilliant staff help me out?

Aditya Parson - 4 years, 5 months ago

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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 - 4 years, 5 months ago

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How did you get this?

Aditya Parson - 4 years, 5 months ago

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How did you get sint in the numerator?

Aditya Parson - 4 years, 5 months ago

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IMPOSSIBLE! No dx anywhere to be found! :D

Kenneth Chan - 4 years, 5 months ago

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

Aditya Parson - 4 years, 5 months ago

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

Riya Gupta - 4 years, 5 months ago

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just write it in fragments, I will understand.

Aditya Parson - 4 years, 5 months ago

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

Aditya Parson - 4 years, 5 months ago

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

will u w8?

Riya Gupta - 4 years, 5 months ago

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@Riya Gupta Yeah I will wait.

Aditya Parson - 4 years, 5 months ago

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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?

Riya Gupta - 4 years, 5 months ago

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@Riya Gupta No that is not the answer.. But I would like to see your working, as to how you derived the above solution.

Aditya Parson - 4 years, 5 months ago

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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 - 4 years, 5 months ago

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

Aditya Parson - 4 years, 5 months ago

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

Riya Gupta - 4 years, 5 months ago

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Sure. No problem.

Aditya Parson - 4 years, 5 months ago

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:)

Riya Gupta - 4 years, 5 months ago

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SIMPLY CLICK HERE

Raja Metronetizen - 4 years, 5 months ago

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The last steps are WRONG!! Check it yourself.

Aditya Parson - 4 years, 5 months ago

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

Raja Metronetizen - 4 years, 5 months ago

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

Aditya Parson - 4 years, 5 months ago

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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...

Raja Metronetizen - 4 years, 5 months ago

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@Raja Metronetizen 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 - 4 years, 5 months ago

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@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 ......

Raja Metronetizen - 4 years, 5 months ago

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@Raja Metronetizen No, I have not yet got the answer.

Aditya Parson - 4 years, 5 months ago

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

Aditya Parson - 4 years, 5 months ago

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Wrong.

Aditya Parson - 4 years, 5 months ago

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Hi,

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

Gopinath No - 4 years, 5 months ago

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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.......:)

Raja Metronetizen - 4 years, 5 months ago

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

Gopinath No - 4 years, 5 months ago

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@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....

Raja Metronetizen - 4 years, 5 months ago

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I think you are missing the limits here.

Gopinath No - 4 years, 5 months ago

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Huh? Indefinite integration!!

Aditya Parson - 4 years, 5 months ago

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what limits has to do with this question?

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

Riya Gupta - 4 years, 5 months ago

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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 - 4 years, 5 months ago

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@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 - 4 years, 5 months ago

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@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 - 4 years, 5 months ago

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@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 - 4 years, 5 months ago

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

Gopinath No - 4 years, 5 months ago

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@Gopinath No RD sharma

Aditya Parson - 4 years, 5 months ago

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

Gopinath No - 4 years, 5 months ago

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

Aditya Parson - 4 years, 5 months ago

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

Riya Gupta - 4 years, 5 months ago

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@Riya Gupta Uhmm.. Pardon me?

Aditya Parson - 4 years, 5 months ago

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

Riya Gupta - 4 years, 5 months ago

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@Riya Gupta 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 - 4 years, 5 months ago

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@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.....

Riya Gupta - 4 years, 5 months ago

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@Riya Gupta I will wait..

Aditya Parson - 4 years, 5 months ago

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

Aditya Parson - 4 years, 5 months ago

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

Riya Gupta - 4 years, 5 months ago

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@Riya Gupta Yeah, I did.

Aditya Parson - 4 years, 5 months ago

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@Aditya Parson ok

Riya Gupta - 4 years, 5 months ago

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

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

Riya Gupta - 4 years, 5 months ago

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