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

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

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## Comments

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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.

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

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I did the same thing yesterday.

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IMPOSSIBLE! No dx anywhere to be found! :D

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

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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 ...

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How did you get sint in the numerator?

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How did you get this?

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No solutions. Can the brilliant staff help me out?

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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....:(

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Sure. No problem.

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:)

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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

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

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

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just write it in fragments, I will understand.

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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?

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

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will u w8?

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SIMPLY CLICK HERE

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Hi,

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

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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.......:)

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Wrong.

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The last steps are WRONG!! Check it yourself.

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

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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...

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I think you are missing the limits here.

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Huh? Indefinite integration!!

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what limits has to do with this question?

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

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

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

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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..

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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.....

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