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

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

No vote yet

3 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

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

Log in to reply

I did the same thing yesterday.

Log in to reply

Log in to reply

No solutions. Can the brilliant staff help me out?

Log in to reply

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

Log in to reply

How did you get this?

Log in to reply

How did you get sint in the numerator?

Log in to reply

IMPOSSIBLE! No dx anywhere to be found! :D

Log in to reply

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???/

Log in to reply

just write it in fragments, I will understand.

Log in to reply

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

Log in to reply

will u w8?

Log in to reply

Log in to reply

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?

Log in to reply

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

Log in to reply

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....:(

Log in to reply

Sure. No problem.

Log in to reply

:)

Log in to reply

SIMPLY CLICK HERE

Log in to reply

The last steps are WRONG!! Check it yourself.

Log in to reply

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

Log in to reply

Log in to reply

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...

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Wrong.

Log in to reply

Hi,

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

Log in to reply

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

Log in to reply

Log in to reply

I think you are missing the limits here.

Log in to reply

Huh? Indefinite integration!!

Log in to reply

what limits has to do with this question?

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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.....

Log in to reply

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply