Evaluate $\int_0^{\pi/2}\ln(\tan^4(x)+1)dx$ I would like to know how to evaluate $\int_0^{\pi/2}\ln(\tan^4(x)+1)dx$
I tried integrating backwards, that is, letting $u = \pi/2  - x$, and this gave me
$$\int_0^{\pi/2}\ln(\tan^4(\pi/2 - u)+1)du = \int_0^{\pi/2}\ln(\cot^4(u)+1)du \\= \int_0^{\pi/2}\ln(\frac{\tan^4(u)+1}{\tan^4(x)})du \\= \int_0^{\pi/2}\ln(\tan^4(u)+1)du - \int_0^{\pi/2}\ln(\tan^4(u))du$$
This tells us that:
$\int_0^{\pi/2}\ln(\tan^4(u))du = 0$
which is great but not what we want.
Any advice would be appreciated. Thank you!
 A: $$I=\int_{0}^{\pi/2} \ln(1+\tan^4 t)dt= \int_{0}^{\pi/2} \ln \frac{\sin^4 t+\cos^4 t}{\cos^4 t} dt$$ $$I =\int_{0}^{\pi/2}  \ln [(\sin^2 t+\cos^2t)^2-2\sin^2 t \cos^2 t] dt-4 \int_{0}^{\pi/2}\ln  \cos t dt$$ $$I=\int_{0}^{\pi/2} \ln (1-\frac{1}{2} \sin^2 2t) dt-4 \int_{0}^{\pi/2} \ln \cos t ~ dt$$ $$I=\int_ {0}^{\pi/2} \ln (3+\cos 4t) dt-\pi \ln 2-4J=\frac{1}{2} \int_{0}^{\pi} \ln (3+ \cos u) du-\pi \ln 2-4J ~~~~(1) $$
Let $$K(a)=\int_{0}^{\pi} \frac{dz}{a+\cos z} dz=\int_{0}^{\pi} \frac{dz}{a+\cos (\pi-z)} dz \implies 2K(a)= \int_{0}^{\pi} \frac{2adz}{a^2-\cos^2 z} dt$$
Use $\tan z=u$ to get
$$K(a)=\frac{1}{a^2}\int_{0}^{\infty} \frac{a}{u^2+(a^2-1)/a^2}=\frac{\pi}{\sqrt{a^2-1}}$$
Integrating w.r.t. $a$ in above we get
$$\int_{0}^{\pi} \ln(a+\cos z) dz = \pi \ln (a+\sqrt{a^2-1})+C~~~(2)$$
Taking $a=1$ in above, we get $C=\int_{0}^{\pi} \ln [2 \cos^2 (z/2)] dz =\pi \ln 2+4J.$
Inserting (2) in (1), for $a=3$, we finally get
$$I=\frac{\pi}{2} \ln[ 2(3+2\sqrt{2})]=\pi \ln(2+\sqrt{2}).$$
Note that: $$J=\int_{0}^{\pi/2} \ln \cos t~dt =-(\pi/2)\ln 2.$$is a well known integral.
A: Note
$$I=\int_0^{\pi/2}\ln(1+\tan^4x)dx\overset{t=\tan x}=\int_0^\infty \frac{\ln(1+t^4)}{1+t^2}dt
$$
Let $J(a) = \int_0^\infty \frac{\ln(1+a^4t^4)}{1+t^2}dt$, along with $J(0)=0$ and
$$J’(a) =\hspace{-2mm} \int_0^\infty\hspace{-3mm} \frac{4a^3t^4\ dt}{(1+a^4t^4)(1+t^2)}
= \frac{4a^3}{1+a^4}\int_0^\infty \left( \frac{1}{1+t^2}
 + \frac{t^2-1}{1+a^4t^4}\right)dt
= \frac\pi2 \frac{4a^3}{1+a^4} +\pi\sqrt2\frac {1-a^2}{1+a^4}
$$
Then
$$I=J(1)=\int_0^1 J’(a)da =\frac\pi2\int_0^1 \frac{4a^3}{1+a^4} da 
 + \pi\sqrt2 \int_0^1 \frac{1-a^2}{1+a^4}da \\
=\frac\pi2\ln2+\pi\coth^{-1}\sqrt2= \pi\ln(2+\sqrt2) $$
