Evaluation of $$\int^{1}_{0}\frac{\ln(13-6x)}{\sqrt{1-x^2}}dx$$
$\bf{My\; Try::}$ Put $x=\cos \theta\;,$ Then $dx = -\sin \theta d\theta$ and changing limits, We get
$$I = -\int^{0}_{\frac{\pi}{2}}\frac{\ln(13-6\cos \theta)}{\sin \theta}\cdot \sin \theta d \theta = \int^{\frac{\pi}{2}}_{0}\ln(13-6\cos \theta) d\theta$$
Now How can i solve it , Help required, Thanks
HINT $$ \int \ln (13-6x) \frac{dx}{\sqrt{1-x^2}} $$ integrates by parts, where you can differentiate the log and integrate the right factor getting an $\arcsin$. The new one should be a trigonometric form.
Let $\sin a=\frac6{13}$ and substitute $x=\cos t$
\begin{align} & \int^{1}_{0}\frac{\ln(13-6x)}{\sqrt{1-x^2}}dx\\ = & \ \frac\pi2\ln13 + \int_0^{\pi/2} \ln(1-\sin a \cos t)dt\\ =& \ \frac\pi2 \ln13 -\int_0^\frac{\pi}{2}\int_0^a \frac{\cos s \cos t}{1-\sin s\cos t}ds \ dt\\ =& \ \frac\pi2 \ln13 -\int_0^a \left(\frac\pi2\tan\frac s2+\frac{s}{\sin s}\right)ds \\ =& \ \frac\pi2 \ln13+\pi \ln\left(\cos\frac {a}2\right)-2 \text{Ti}_2\left(\tan\frac a2 \right)\\ =&\ \frac\pi2\ln\frac{13+\sqrt{133}}{2}-2\text{Ti}_2\bigg(\frac{13-\sqrt{133}}{6}\bigg) \end{align} where $ \int_0^a\frac{s}{\sin s}ds=2\text{Ti}_2(\tan\frac a2 )$.
Use Euler substitution $$u=\frac{1-\sqrt{1-x^2}}{x}$$ $$I=\int_0^1\frac{\ln(13-6x)}{\sqrt{1-x^2}}dx=2\int_0^1\ln\left(\frac{13u^2-12u+13}{1+u^2}\right)\frac{du}{1+u^2}$$$$=2\int_0^1\frac{\ln13}{1+u^2}du+2\int_0^1\frac{\ln(u^2-\frac{12}{13}u+1)}{1+u^2}du-2\int_0^1\frac{\ln(1+u^2)}{1+u^2}du$$ $$=\frac{\pi\ln(13/4)}{2}+2G+2\int_0^1\frac{\ln(u^2-\frac{12}{13}u+1)}{1+u^2}du$$Using the following formula, the last integral can be expressed using dilogarithms. $r$ represents a root of $u^2-\frac{12}{13}u+1$ $$J(a,b)=\int_0^1\frac{\ln(x+a)}{x+b}dx=\ln(a-b)\ln\left(1+\frac{1}{b}\right)+\operatorname{Li}_2\left(\frac{b}{b-a}\right)-\operatorname{Li}_2\left(\frac{1+b}{b-a}\right)$$ $$I=\frac{\pi\ln(13/4)}{2}+2G-2\Im\int_0^1\frac{\ln((u-r)(u-1/r))}{u+i}du$$ $$I=\frac{\pi\ln(13/4)}{2}+2G-2\Im\left[J(-r,i)-J\left(-\frac{1}{r},i\right)\right]$$
Substitute $u=13-6x$ and $\text{d}u=-6\space\text{d}x$
$$\mathcal{I}=\int_0^1\frac{\ln\left(13-6x\right)}{\sqrt{1-x^2}}\space\text{d}x=-\frac{1}{6}\int_{13}^7\frac{\ln\left(u\right)}{\sqrt{1-\left(\frac{13-u}{6}\right)^2}}\space\text{d}u=-\int_{13}^7\frac{\ln\left(u\right)}{\sqrt{36-(13-u)^2}}\space\text{d}u$$
Now, using integration by parts:
$$\mathcal{I}=\left[\ln(u)\arcsin\left(\frac{13-u}{6}\right)\right]_{13}^7-\int_{13}^7\frac{\arcsin\left(\frac{13-u}{6}\right)}{u}\space\text{d}u$$
Use:
$$\ln(7)\arcsin\left(\frac{13-7}{6}\right)-\ln(13)\arcsin\left(\frac{13-13}{6}\right)=\frac{\pi\ln(7)}{2}$$
$$I=\int^{\frac{\pi}{2}}_{0}\ln(13-6\sin \theta) d\theta =\frac\pi 2\ln13+\int^{\frac{\pi}{2}}_{0}\ln(1-\frac6{13}\sin \theta) d\theta$$ By using the expansion $\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$, $$I=\frac\pi 2\ln13-\sum_{n=1}^\infty\frac{(\tfrac6{13})^n}n\int^{\frac{\pi}{2}}_{0}\sin^n\theta d\theta$$ $$I=\frac\pi 2\ln13-\sum_{n=1}^\infty\frac{B(\frac{n+1}2,\frac12)}{2n}(\tfrac6{13})^n$$ where $B(p,q)=2\int^{\frac{\pi}{2}}_{0}\sin^{2p-1}\theta\cos^{2q-1}\theta d\theta$ is the beta function.
If $n=2m$, $B(\frac{n+1}2,\frac12)=\frac{\Gamma(\frac{2m+1}2)\Gamma(\frac12)}{\Gamma(\frac{2m+1}2+\frac12)}=\frac{\Gamma(\frac{2m+1}2)\pi}{\Gamma(\frac12)\Gamma(m+1)}=\pi\frac{(\tfrac12)_m}{(1)_m}=\pi c_m$ where $c_m={2m\choose m}2^{-2m}$ as defined in this answer and $(a)_m=\frac{\Gamma(m+a)}{\Gamma(a)}$ is the Pochammer symbol.
If $n=2m+1$, $B(\frac{n+1}2,\frac12)=\frac{\Gamma(\frac{2m+2}2)\Gamma(\frac12)}{\Gamma(\frac{2m+2}2+\frac12)}=\frac{\Gamma(m+1)\Gamma(\frac12)}{\frac{2m+1}2\Gamma(m+\frac12)}=\frac{2(1)_m}{(2m+1)(\tfrac12)_m}=\frac2{(2m+1)c_m}.$
The sum in $I$ splits and we have $I=\frac\pi 2\ln13-S_1-S_2$ where $S_1=\frac\pi 4\sum_{m=1}^\infty\frac{c_m}{m}(\tfrac6{13})^{2m}$ and $S_2=\sum_{m=0}^\infty\frac{1}{(2m+1)^2c_m}(\tfrac6{13})^{2m+1}.$
With similar ideas in this answer, we can show that $$\sum_{m=1}^\infty\frac{c_m}{m}x^m=\ln4-2\ln(1+\sqrt{1-x}).$$ By plugging $x=\frac{36}{169}$ we have $S_1=\frac\pi 2\ln13-\frac\pi 2\ln\frac{13+\sqrt{133}}2.$
Computation of $S_2$: Since $\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{m=0}^\infty \frac1{(2m+1)c_m}x^{2n+1}$, $$S_2=\int_0^{\frac{6}{13}}\frac{\arcsin x}{x\sqrt{1-x^2}}dx,$$ $$S_2=\int_0^{\arcsin\frac{6}{13}}\frac{\theta}{\sin\theta}d\theta.$$
Thus, $$I=\frac\pi 2\ln\frac{13+\sqrt{133}}2-S_2.$$
