I need help in solving the following problem:
$$\int \ln\left(\sin(\sqrt{x})+\cos(\sqrt{x})\right)dx $$
I really don't know how to start solving this problem; any tips or solutions will be greatly appreciated.
Thank you in advance.
I need help in solving the following problem:
$$\int \ln\left(\sin(\sqrt{x})+\cos(\sqrt{x})\right)dx $$
I really don't know how to start solving this problem; any tips or solutions will be greatly appreciated.
Thank you in advance.
On
I really don't know how to start solving this problem
This is hardly surprising, since the anti-derivative cannot be expressed in terms of elementary functions.
any tips or solutions will be greatly appreciated.
Add some integration limits:
$$\int_0^{\big(\tfrac\pi2\big)^2}\ln\Big(\sin\big(\sqrt{x}\big)+\cos\big(\sqrt{x}\big)\Big)~dx ~=~ \frac\pi2~\bigg(\text{Catalan}-\frac\pi4~\ln2\bigg)$$
$$\int_0^{\big(\tfrac\pi4\big)^2}\ln\Big(\sin\big(\sqrt{x}\big)+\cos\big(\sqrt{x}\big)\Big)~dx ~=~ \frac{21~\zeta(3)-2\pi^2\ln2}{64}$$
By setting $x=u^2$ we have: $$ I = \int 2u\log(\sin u+\cos u)\,du = u^2\log(\sin u+\cos u)-\int u^2\frac{1-\tan u}{1+\tan u}\,du$$ and, by putting $v=\frac{\pi}{4}-v$, $$ -\int u^2\frac{1-\tan u}{1+\tan u}\,du = \int \left(\frac{\pi}{4}-v\right)^2 \tan v\,dv $$ Now we may exploit $\int\tan v\,dv=-\log\cos v $, so the last integral just depends on: $$ \int v \log(\cos v)\,dv $$ that, however, is not an elementary function, but a combination of a logarithm, a dilogarithm and a trilogarithm multiplied by powers of $v$: just write $\cos v$ as $\frac{e^{iv}+e^{-iv}}{2}$, exploit the Taylor series of $\log(1+z)$ and integrate termwise.