# Integrate this.

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

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

5 years, 8 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.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

I did the same thing yesterday.

- 5 years, 8 months ago

go to that post i posted ysterdae....

- 5 years, 8 months ago

IMPOSSIBLE! No dx anywhere to be found! :D

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

How did you get sint in the numerator?

- 5 years, 8 months ago

How did you get this?

- 5 years, 8 months ago

No solutions. Can the brilliant staff help me out?

- 5 years, 8 months ago

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

- 5 years, 8 months ago

Sure. No problem.

- 5 years, 8 months ago

:)

- 5 years, 8 months 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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

just write it in fragments, I will understand.

- 5 years, 8 months ago

i got

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

will u w8?

- 5 years, 8 months ago

Yeah I will wait.

- 5 years, 8 months ago

- 5 years, 8 months ago

Hi,

- 5 years, 8 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.......:)

- 5 years, 8 months ago

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

- 5 years, 8 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....

- 5 years, 8 months ago

Wrong.

- 5 years, 8 months ago

The last steps are WRONG!! Check it yourself.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months 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...

- 5 years, 8 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 : $$[1-(sinx-cosx)^2]^2$$

- 5 years, 8 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 ......

- 5 years, 8 months ago

No, I have not yet got the answer.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

I think you are missing the limits here.

- 5 years, 8 months ago

Huh? Indefinite integration!!

- 5 years, 8 months ago

what limits has to do with this question?

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

yeah thats what i thought.................

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

- 5 years, 8 months ago

u edited the question few hours back????????

- 5 years, 8 months ago

Yeah, I did.

- 5 years, 8 months ago

ok

- 5 years, 8 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..

- 5 years, 8 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.

- 5 years, 8 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.

- 5 years, 8 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.

- 5 years, 8 months ago

- 5 years, 8 months ago

RD sharma

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

Uhmm.. Pardon me?

- 5 years, 8 months ago

sorry i didnt got what do you mean?

- 5 years, 8 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?

- 5 years, 8 months ago

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.....

- 5 years, 8 months ago

I will wait..

- 5 years, 8 months ago

And yeah the latest edition it is...

- 5 years, 8 months ago