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

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

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

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

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

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

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

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, 3 months ago

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– Riya Gupta · 4 years, 3 months ago

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

– Aditya Parson · 4 years, 3 months ago

I did the same thing yesterday.Log in to reply

– Riya Gupta · 4 years, 3 months ago

go to that post i posted ysterdae....Log in to reply

No solutions. Can the brilliant staff help me out? – Aditya Parson · 4 years, 3 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, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

How did you get this?Log in to reply

– Aditya Parson · 4 years, 3 months ago

How did you get sint in the numerator?Log in to reply

IMPOSSIBLE! No dx anywhere to be found! :D – Kenneth Chan · 4 years, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

Now it is there.. Thanks for pointing it out!Log in to reply

I got an answer......but dont know how to write it in latex.....what to do???/ – Riya Gupta · 4 years, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

just write it in fragments, I will understand.Log in to reply

– Aditya Parson · 4 years, 3 months ago

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

will u w8? – Riya Gupta · 4 years, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

Yeah I will wait.Log in to reply

-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, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

No that is not the answer.. But I would like to see your working, as to how you derived the above solution.Log in to reply

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, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

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

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, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

Sure. No problem.Log in to reply

– Riya Gupta · 4 years, 3 months ago

:)Log in to reply

SIMPLY CLICK HERE – Raja Metronetizen · 4 years, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

The last steps are WRONG!! Check it yourself.Log in to reply

– Raja Metronetizen · 4 years, 3 months ago

oh...sorry...sorry....i mistook doing hastily...Log in to reply

– Aditya Parson · 4 years, 3 months ago

Well, the link you provide is too clumsy. I would appreciate it if someone posts the solution with precision.Log in to reply

solution..........HOPE THIS HELPS..............THE FINAL STEP,YOU CAN SEE,HERE........THANX... – Raja Metronetizen · 4 years, 3 months ago

here is the full solution......please let me know if there is any wrong in thisLog in to reply

– Aditya Parson · 4 years, 3 months ago

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 : \( [1-(sinx-cosx)^2]^2\)Log in to reply

– Raja Metronetizen · 4 years, 3 months ago

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 ......Log in to reply

– Aditya Parson · 4 years, 3 months ago

No, I have not yet got the answer.Log in to reply

– Aditya Parson · 4 years, 3 months ago

I already knew these steps you put up here, but they did not take me anywhere close to the answer.Log in to reply

– Aditya Parson · 4 years, 3 months ago

Wrong.Log in to reply

How did you get the step before your answer? Did you try differentiating the answer? – Gopinath No · 4 years, 3 months ago

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– Raja Metronetizen · 4 years, 3 months 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.......:)Log in to reply

– Gopinath No · 4 years, 3 months ago

Text books can be wrong. What I meant was differentiating your answer and see if it matches with the question.Log in to reply

– Raja Metronetizen · 4 years, 3 months ago

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....Log in to reply

I think you are missing the limits here. – Gopinath No · 4 years, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

Huh? Indefinite integration!!Log in to reply

m getting confused........... – Riya Gupta · 4 years, 3 months ago

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– Gopinath No · 4 years, 3 months ago

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

– Aditya Parson · 4 years, 3 months ago

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.Log in to reply

– Gopinath No · 4 years, 3 months ago

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.Log in to reply

– Aditya Parson · 4 years, 3 months ago

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.Log in to reply

– Gopinath No · 4 years, 3 months ago

Your textbook has an answer? What's the name of the book?Log in to reply

– Aditya Parson · 4 years, 3 months ago

RD sharmaLog in to reply

– Gopinath No · 4 years, 3 months ago

Okay, I had that book once upon a time. Is it the latest edition ?Log in to reply

– Aditya Parson · 4 years, 3 months ago

And yeah the latest edition it is...Log in to reply

– Riya Gupta · 4 years, 3 months ago

the same edition you are having can work...[if u r still having R.D. ]Log in to reply

– Aditya Parson · 4 years, 3 months ago

Uhmm.. Pardon me?Log in to reply

– Riya Gupta · 4 years, 3 months ago

sorry i didnt got what do you mean?Log in to reply

– Aditya Parson · 4 years, 3 months ago

Sorry, I mistook your previous comment as directed towards me, so I'd for a moment lost context. Anyways, no solutions for this integral?Log in to reply

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, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

I will wait..Log in to reply

– Aditya Parson · 4 years, 3 months ago

It has nothing to with limits. It is indefinite integration.Log in to reply

– Riya Gupta · 4 years, 3 months ago

u edited the question few hours back????????Log in to reply

– Aditya Parson · 4 years, 3 months ago

Yeah, I did.Log in to reply

– Riya Gupta · 4 years, 3 months ago

okLog in to reply

thanks for clearing my doubt........:) – Riya Gupta · 4 years, 3 months ago

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