Mostrar que $\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi$

Question

Dado que:

$$\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi\tag1$$

$$1-\sin^2(x)+\sin^2(x)\tan^4(x)$$ $$=1+\sin^2(x)[\tan^4(x)-1]$$ $$=1+\sin^2(x)[\tan^2(x)-1][\tan^2(x)+1]$$ $$=1+\tan^2(x)[\tan^2(x)-1]$$ $$=1-\tan^2(x)+\tan^4(x)$$

$$\int_{0}^{\pi/2}{\ln[1-\tan^2(x)+\tan^4(x)]\over \sin^2(x)}\mathrm dx\tag2$$

$t=\tan(x)\implies dt=\sec^2(x) dx$

$$\int_{0}^{\infty}{\ln(1-t^2+t^4)\over t^2}\mathrm dt\tag3$$

$u=t^2\implies du=2tdt$

$${1\over 2}\int_{0}^{\infty}{\ln(1-u+u^2)\over u^{3/2}}\mathrm du\tag4$$

Answer 1

Usted está casi allí. Usted puede notar que $1-t^2+t^4 = \frac{1+t^6}{1+t^2}$ y

$$ \int_{0}^{+\infty}\frac{\log(1+t^2)}{t^2}\,dt = \pi, \qquad \int_{0}^{+\infty}\frac{\log(1+t^6)}{t^2}\,dt = 2\pi \tag{1} $$ son bastante fáciles de probar. Ambos dependen de $$ \int_{0}^{+\infty}\frac{\log(1+t)}{t^\alpha}\,dt \stackrel{\text{IBP}}{=} \frac{\pi}{(1-\alpha)\sin(\pi\alpha)}.\tag{2}$$

Answer 2

$J=\displaystyle \int_{0}^{\infty}{\ln(1-t^2+t^4)\over t^2}\mathrm dt$

$\begin{align} J&=\Big[-\dfrac{\ln(1-t^2+t^4)}{t}\Big]_{0}^{+\infty}+\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt\\ &=\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt \end{align}$

Observar que,

$\displaystyle\int_{0}^{+\infty}\dfrac{t^2}{1-t^2+t^4}dt=\int_{0}^{+\infty}\dfrac{1}{1-t^2+t^4}dt$

(realizar el cambio de variable $y=\dfrac{1}{t}$ )

Por lo tanto,

$\begin{align} J&=\int_{0}^{+\infty}\dfrac{2}{1-t^2+t^4}dt\\ &=\int_{0}^{+\infty}\dfrac{t^2}{1-t^2+t^4}dt+\int_{0}^{+\infty}\dfrac{1}{1-t^2+t^4}dt\\ &=\int_{0}^{+\infty}\left(1+\dfrac{1}{t^2}\right)\dfrac{1}{1+\left(t-\tfrac{1}{t}\right)^2}dt \end{align}$

Realizar el cambio de variable $y=t-\dfrac{1}{t}$,

$\begin{align} J&=\int_{-\infty}^{+\infty}\dfrac{1}{1+t^2}dt\\ &=\boxed{\pi} \end{align}$