Ytukyg

As part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:

$$ int_{0}^{frac{pi}{2}} lnleft|2 + tan^2(x) right| :dx $$

I'm seeking methods using the Feynman Trick (or any method for that matter) that can be used to solve this definite integral.

If one wishes to use "Feynman's Trick," then begin by defining a function $I(a)$, $a>1$ as given by

$$I(a)=int_0^{pi/2}log(a+tan^2(x)),dx tag1$$

Differentiation of $(1)$ reveals

$$begin{align}
I'(a)&=int_0^{pi/2} frac{1}{a+tan^2(x)},dx\\
&=frac{pi/2}{a-1}-frac{pi/2}{sqrt a (a-1)}tag2
end{align}$$

Integration of $(2)$ yields

$$begin{align}
I(a)&=fracpi2left(log(a-1)+logleft(frac{sqrt a+1}{sqrt{a}-1}right) right)\\
&=pi log(sqrt a+1)tag3
end{align}$$

Finally, setting $a=2$ in $(3)$, we obtain the coveted result

$$int_0^{pi/2}log(2+tan^2(x)),dx=pi log(sqrt 2+1)$$

My approach

Let

begin{align}
int_{0}^{frac{pi}{2}} lnleft|2 + tan^2(x) right| :dx &= int_{0}^{frac{pi}{2}} lnleft|1 + left(1 + tan^2(x)right) right| :dx \
&= int_{0}^{frac{pi}{2}} lnleft|1 + sec^2(x) right| :dx \
&= int_{0}^{frac{pi}{2}} lnleft|frac{cos^2(x) + 1}{cos^2(x)} right| :dx \
&= int_{0}^{frac{pi}{2}} left[ lnleft|cos^2(x) + 1 right| - lnleft|cos^2(x)right| right]:dx \
&= int_{0}^{frac{pi}{2}} lnleft|cos^2(x) + 1 right|:dx - int_{0}^{frac{pi}{2}} lnleft|cos^2(x)right|:dx
end{align}

Now

$$ int_{0}^{frac{pi}{2}} lnleft|cos^2(x)right|:dx = 2int_{0}^{frac{pi}{2}} lnleft|cos(x)right|:dx = 2cdot-frac{pi}{2}ln(2) = -pi ln(2)$$

For detail on this definite integral, see guidance here

We now need to solve

$$ int_{0}^{frac{pi}{2}} lnleft|cos^2(x) + 1 right|:dx $$

Here, Let

$$ I(t) = int_{0}^{frac{pi}{2}} lnleft|cos^2(x) + t right|:dx $$

Thus,

$$
frac{dI}{dt} = int_{0}^{frac{pi}{2}} frac{1}{cos^2(x) + t}:dx
= int_{0}^{frac{pi}{2}} frac{1}{frac{cos(2x) + 1}{2} + t}:dx = 2int_{0}^{frac{pi}{2}} frac{1}{cos(2x) + 2t + 1}:dx
$$

Employ a change of variable $u = 2x$:

$$frac{dI}{dt} = int_{0}^{pi} frac{1}{cos(u) + 2t + 1}:du $$

Employ the Weierstrass substitution $omega = tanleft(frac{u}{2} right)$:

begin{align}
frac{dI}{dt} &= int_{0}^{infty} frac{1}{frac{1 - omega^2}{1 + omega^2} + 2t + 1}:frac{2}{1 + omega^2}cdot domega \
&= int_{0}^{infty} frac{1}{tomega^2 + t + 1} :domega \
&= frac{1}{t}int_{0}^{infty} frac{1}{omega^2 + frac{t + 1}{t}} :domega \
&= frac{1}{t}left[frac{1}{sqrt{frac{t+1}{t}}}arctanleft( frac{omega}{sqrt{frac{t+1}{t}}}right)right]_{0}^{infty} \
&= frac{1}{t}frac{1}{sqrt{frac{t+1}{t}}}frac{pi}{2} \
&= frac{1}{sqrt{t}sqrt{t + 1}}frac{pi}{2}
end{align}

And so,

$$I(t) = int frac{1}{sqrt{t}sqrt{t + 1}}frac{pi}{2}:dt = pilnleft| sqrt{t} + sqrt{t + 1}right| + C$$

Now

$$I(0) = int_{0}^{frac{pi}{2}} lnleft|cos^2(x) + 0 right|:dx = -pi ln(2) = pilnleft|sqrt{0} + sqrt{0 + 1} right| + C rightarrow C = -pi ln(2)$$

And so,

$$I(t) = pilnleft| sqrt{t} + sqrt{t + 1}right| -pi ln(2) = pilnleft|frac{sqrt{t} + sqrt{t + 1}}{2} right|$$

Thus,

$$I = I(1) = int_{0}^{frac{pi}{2}} lnleft|cos^2(x) + 1 right|:dx = pilnleft|frac{sqrt{1} + sqrt{1 + 1}}{2} right| = pilnleft|frac{1 + sqrt{2}}{2} right| $$

And Finally

begin{align}
int_{0}^{frac{pi}{2}} lnleft|2 + tan^2(x) right| :dx &= int_{0}^{frac{pi}{2}} lnleft|cos^2(x) + 1 right|:dx - int_{0}^{frac{pi}{2}} lnleft|cos^2(x)right|:dx \
&= pilnleft|frac{1 + sqrt{2}}{2} right| - left(-pi ln(2)right) \
&= pilnleft|1 + sqrt{2} right|
end{align}

Good answers have already been posted. Maybe one will like this way too: $$I=int_0^frac{pi}{2}ln(2+tan^2x)dxoverset{tan x=t}=int_0^infty frac{ln(2+t^2)}{1+t^2}dt$$
We now consider the following integral: $$I(a)= int_0^infty frac{ln(1+a(1+x^2))}{1+x^2}dxrightarrow I'(a)=int_0^infty frac{1+x^2}{(1+a(1+x^2))(1+x^2)}dx$$
$$=int_0^infty frac{dx}{1+a+ax^2}= frac{1}{sqrt{a+a^2}} arctanleft(sqrt{frac{ a} {a+1}}xright)bigg|_0^infty=frac{pi}{2}frac{1}{sqrt{a+a^2}}$$
Since $I(0)=0$ We have that $I=I(1)-I(0)= int_0^1 I'(a)da $
$$I=frac{pi}{2} int_0^1 frac{da}{sqrt{a}sqrt{1+a}} overset{sqrt a=t}=piint_0^1 frac{dt}{sqrt{1+t^2}}=piln(1+sqrt 2)$$