Could anybody help me to solve this, a question for brilliant guys and girls?

Question

integrate sinx + cosx by sin to the power 4x + cos to the power 4x

∫SinX+CosX / Sin(To the Power 4)X + Cos (To the Power 4) X = ?

This is a question which is not getting solved in my entire class. Pls help me out.

∫(SinX+CosX / Sin(To the Power 4)X + Cos (To the Power 4) X)dX = ?

Sorry for mistake.

Answer 1

first of all let me tell u tht 'to the power'is reprsented by ^. so i will be using this

now I=(sinx+cosx)/(sin^4x+cos^4x)
write sin^4x+cos^4x= (sin^2x+cos^2x)^2 - 2sin^2x.cos^2x
= [ (sinx+cosx)^2 - 2sinx.cosx]^2 - 2sin^2x.cos^2x

put this value if sin^4x+cos^4x in I
now the trick is put sinx-cosx=t
so diffrentiating
cosx+sinx=dt/dx
dx=dt/(sinx+cosx)

also on squaring sinx-cosx=t we get
1-2sinx.cosx=t^2
sinx.cosx=(1-t^2)/2

now put the value of dx and sinx.cosx in I and thus u will be able to get ur answer.

if u r wondering abt the trick then the rule is tht
if there is sinx+cosx in numerator then put sinx-cosx=t
if there is sinx-cosx in the numerator then put sinx+cosx=t


hope u got ur answer:-))

Answer 2

Do you mean

INT (sin x + cos x) / (sin^4 x + cos^4 x) dx ?

This integral is the sum of two integrals; if we write u = sin x, v = cos x, then it is

INT du / (1 - 2u^2 + 2u^4) - INT dv / (1 - 2v^2 + 2v^4)

The denominator (1 - 2u^2 + 2u^4) can be factored as

(1 - K u + H u^2) (1 + K u + H u^2)

with H = sqrt(2), K = sqrt(2 + 2H)

If we take J = H/2K then

1 / (1 - 2 u^2 + 2 u^4) =
... = (J u - 1/2) / (1 - K u + H u^2)
... + (J u + 1/2) / (1 + K u + H u^2)

I focus on the first term; the second is similar:

Write E = J/(2H), F = K*E, and G = 1/2 - F. Then (J u - 1/2) = E*(2H u - K) - G, and the factor (2H u - K) is the derivative of the denominator. Therefore,

INT (J u - 1/2) / (1 - K u + H u^2) du =
... = E ln |1 - Ku + Hu^2|
... - G INT du / (1 - Ku + Hu^2)

For the last integral, we calculate the discriminant, D = 4H - K^2 > 0, and take the square root q = sqrt(D). Now

INT du / (1 - Ku + Hu^2) = (2/q) * inv tan {(2H u - K) / q}

so, with A = 2G/q, B = 2H/q and C = K/q,

INT (J u - 1/2) / (1 - K u + H u^2) du =
... = E ln |1 - Ku + Hu^2|
... - A * inv tan (B u - C)

In the same way,
INT (J u + 1/2) / (1 + K u + H u^2) du =
... = E ln |1 + Ku + Hu^2|
... + A * inv tan (B u + C)

Adding these together we find
INT cos x dx / (sin^4 x + cos^4) x =
... = E (ln |1 - Ku + Hu^2| + ln |1 + Ku + Hu2|)
... + A (inv tan (Bu + C) - inv tan (Bu - C))

where u = sin x.

For the sine, take v = cos x and repeat the process -- but take the opposite of the result.

Answer 3

in the beginning, the factorial in line with se is defined purely on integers (for this reason discontinuous), however the gamma function, which evaluates to factorials for integer values, is non-quit and differentiable. For any selection n, its fee is: confident crucial from 0 to infinity of t^(n-a million) cases exp (-t) dt, the place t is a dummy variable. (isn't ASCII astounding for arithmetic? not!) you will desire to be waiting to distinguish under the crucial sign to artwork out the gory information. As for the 2d project, if I even have examine properly what you assert, i don't think of that it does recommendations-set 0. enable n = a million, and play with it slightly and notice what takes place.

Answer 4

Thats a BS question

Answer 5

I honestly have no clue what you mean, try and retype it (your expression) using parenthesis and ^ to express exponetials. It is not an equation, so you can't "solve" it.

Answer 6

Umm, 1?

Answer 7

thats because you teacher sucks

Answer 8

i graphed first the figured it out...i got .23445481

Answer 9

do you have an integral table?

Answer 10

no dx?
and what do you mean by the word "by"?
retype please, as what you have here is unintelligible