# Closed

I have just seen two active posts about integrals of inverse trigonometric function, \$arctan(x)\$, here on MSE. So I decide lớn post this question. This integral comes from a friend of mine (it"s not a homework problem) và we have tried khổng lồ 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 và 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.

Bạn đang xem: Closed  \$ 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 & 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 integrate with respect lớn \$a\$. (For the first term, just substitute \$sqrt a = u\$ và 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 và tedious simplification (see Appendix khổng lồ 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 to lớn 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 to lớn any mistakes, please point it out to lớn 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 and denominator by a certain factor lớn get rid of the fourth roots. For the second, multiply by another factor to 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, and use ccevents.vnematica khổng lồ simplify the kết thúc result.