Solving Trigonometric Equations

     
I have just seen two active posts about integrals of inverse trigonometric function, $arctan(x)$, here on MSE. So I decide to lớn post this question. This integral comes from a friend of mine (it"s not a homework problem) và we have tried lớn evaluate it but no success so far. I have discussed it in chatroom with
Chris"ssis but she gave me a horrible closed-form without proof. You may have a look here & here. My friend doesn"t know the closed-form either. Here is the problem:

$$int_0^pi/2fracsin^2xarctanleft(cos^2x ight)sin^4x+cos^4x,dx$$

Any idea? Any help would be appreciated. Thanks in advance.

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$ ewcommandal<1>eginalign#1endalign enewcommandImoperatornameIm$Result:

$$I = fracpi left<2 pi +log left(-4 sqrt17+13 sqrt2+8 sqrt2+13 ight)-4 an ^-1left(sqrtfrac1sqrt2-frac12+frac1sqrt2+1 ight) ight>8 sqrt2 \approx 0.299397.$$

The evaluation of this integral by hand is not as tedious as I thought at first. It turns out I had already considered this integral last week, in a different form.

First substitute $ an x = t$, as suggested by FDP in the comments. This gives$$I = int_0^infty dt fract^2 arctan left( frac11+t^2 ight)1+t^4.$$Now observe that$$alarctan left( frac11+t^2 ight) &= Im log(i + 1 + t^2) = Im left \&= fracpi4 + Im logleft(1 + frac1-i2 t^2 ight).$$The integral splits up into a trivial part và a less trivial part. We will now find the latter.

Consider the integral$$J(a) = int_0^infty dx fracx^21+x^4 log(1+a x^2).$$We have $J(0) = 0$, and$$alJ"(a) &= int_0^infty dx fracx^4(1+x^4)(1+a x^2)\&= frac11+a^2 int_0^infty dx left< frac11+ a x^2 + fraca x^2 - 11+x^4 ight>\&= fracpi /21+a^2left.$$This is straightforward to lớn integrate with respect to $a$. (For the first term, just substitute $sqrt a = u$ & use partial fractions.)

The result is$$J(a) = fracpi2 sqrt 2 left< log left(a+sqrt2 sqrta+1 ight)+2 arctanleft(sqrt2 sqrta+1 ight) ight>$$After plugging in the limit and tedious simplification (see Appendix to this answer), we obtain$$Im Jleft(frac1-i2 ight) = -fracpi2 sqrt2left\operatornamearccothleft + arctanleft ight$$

Putting everything together,$$alI &= frac pi 4 int_0^infty dt fract^21+t^4 + Im int_0^infty dt fract^2 logleft(1 + frac1-i2 t^2 ight)1+t^4\&= fracpi^28 sqrt 2 + Im Jleft(frac1-i2 ight)\&= fracpi2 sqrt2left frac pi 4-operatornamearccothleft - arctanleft ight. $$

This is numerically equal khổng lồ the claimed result, which I obtained via a completely different route.

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Throughout I have not worried explicitly about choosing the correct branch of $log$. If this led khổng lồ any mistakes, please point it out to me.

AppendixHere I sketch how $Im Jleft(frac1-i2 ight)$ can be calculated. For example, consider the term$$Im logleft(1 + frac1-i2 + sqrt 2 sqrtfrac1-i2 ight) = Im logleft(frac 3 2 - frac i 2 + 2^1/4left(cos frac pi 8 - i sin frac pi 8 ight) ight).$$Using the half-angle formulas for sine & cosine, we can write it as$$al-arctan left(frac1+ sqrt<4>2 sqrt2-sqrt23 + sqrt<4>2 sqrt2+sqrt2 ight) &= - arctanleft(frac2 sqrt5 sqrt2-7+17-2 sqrt2 ight)\&= - arctanleft.$$To obtain the first equality, multiply numerator & denominator by a certain factor to lớn get rid of the fourth roots. For the second, multiply by another factor lớn get rid of all the roots. Of course this simplification is just for aesthetics.

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Moral: introduce a parameter in such a way that the integral with respect lớn the parameter becomes simple, và use gamize.vnematica to simplify the kết thúc result.