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.
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:
$$\ln(7)\arcsin\left(\frac{13-7}{6}\right)-\ln(13)\arcsin\left(\frac{13-13}{6}\right)=\frac{\pi\ln(7)}{2}$$
