1i think using property for a function even about x=0 and using a substitution cos x=t might just solve it!
Evaluate :
\int_{-1}^{1}{\frac{\ln(13-6\cos{x})dx}{\sqrt{1-x^2}}}
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4 Answers
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1708I Think Right Question Should be
\mathbf{\int_{ - 1}^1 {\frac{{\ln \left( {13 - 6x} \right)}}{{\sqrt {1 - {x^2}} }}dx}}.\\\\ $\textbf{So that We can Put $\mathbf{x=cos\;t}$ and $\mathbf{dx=-sin\;t\; dt}$}\\\\ \textbf{The Given Integral Convert into $\mathbf{\int_{0}^{\pi}ln\left(13-6.cos\; t\right)dt}$}
1708$\textbf{First We will Prove that $\mathbf{\int_{0}^{\pi}ln\left(a+b\;cosx\right)dx=\pi.ln\left(\frac{a+\sqrt{a^2-b^2}}{2}\right)}$}$\\\\\\ $\textbf{Now Let $\mathbf{I=\int_{0}^{\pi}ln\left(a+b\;cosx\right)dx}$}$\\\\ $\textbf{Diff. both side w.r.to $\mathbf{b},$ We get$}$\\\\ $\mathbf{\frac{d\;I}{d\;b}=\int_{0}^{\pi}\frac{cos\;x}{a+b\;cos\;x}d\;x}$\\\\\\ $\mathbf{\frac{d\;I}{d\;b}=\frac{1}{b}.\int_{0}^{\pi}\frac{a+b\;cos\;x-a}{a+b\;cos\;x}d\;x=\frac{\pi}{b}-\frac{a}{b}.\int_{0}^{\pi}\frac{1}{a+b\;cos\;x}d\;x$}$\\\\ $\textbf{Now We Will Calculate $\mathbf{\frac{a}{b}.\int_{0}^{\pi}\frac{1}{a+b\;cos\;x}d\;x=\frac{0}{\infty}\frac{2.dt}{(a+b)+(a-b)t^2}dt=\frac{\pi}{\sqrt{a^2-b^2}}}$}$\\\\ $\textbf{By Using the Substitution $\mathbf{tan\left(\frac{x}{2}\right)=t}$}$\\\\ $\textbf{Now Integral $\mathbf{\frac{d\;I}{d\;b}=\frac{\pi}{b}-\frac{\pi.a}{b.\sqrt{a^2-b^2}}}$}$\\\\\\ \mathbf{I=\pi.ln(b)-a\pi\int\frac{d\;b}{b.\sqrt{a^2-b^2}}}$\\\\
$\textbf{Now $\mathbf{\int\frac{d\;b}{b.\sqrt{a^2-b^2}}=\frac{1}{a}.ln(b)-\frac{1}{a}.ln\left(a+\sqrt{a^2-b^2}\right)}$}$\\\\\\ $\textbf{So $\mathbf{I=\pi.ln(b)-\pi.ln(b)+\pi.ln\left(a+\sqrt{a^2-b^2}\right)+C=\pi.ln\left(a+\sqrt{a^2-b^2}\right)+C}$}$\\\\\\ \mathbf{I=\pi.ln\left(a+\sqrt{a^2-b^2}\right)+C.................................(1)}$\\\\\\ $\textbf{Now We will Calculate value of C, Put b=0, Get}$\\\\ $\mathbf{I=\pi.ln(2a)+C}$\\\\ $\mathbf{I=\pi.ln(a)+\pi.ln(2)+C......................................(2)}$\\\\\\ \textbf{Now from $\mathbf{I=\int_{0}^{\pi}ln\left(a+b\;cosx\right)dx}}$\\\\ \textbf{Put b=0,We Get $\mathbf{I=\pi.ln(a)....................................(3)}$}$\\\\\\ $\textbf{From eq..(2) and eq..(3), We Get $\mathbf{C=-\pi.ln(a)}$ }$\\\\ $\textbf{Put Value of C in eqn..(1), We Get}$\\\\\\ \mathbf{I=\pi.ln\left(a+\sqrt{a^2-b^2}\right)-\pi.ln(a)=\pi.ln\left(\frac{a+\sqrt{a^2-b^2}}{a}\right)}$\\\\\\
$\textbf{ So $\mathbf{\int_{0}^{\pi}ln\left(a+b.cos\;x\right)dx=\pi.ln\left(\frac{a+\sqrt{a^2-b^2}}{a}\right)}$ }$\\\\\\ $\textbf{now Put $\mathbf{a=13,b=-6}$, We Get }$\\\\\\ \boxed{\boxed{\mathbf{\int_{0}^{\pi}ln\left(13-6.cos\;x\right)dx=\pi.ln\left(\frac{13+\sqrt{133}}{2}\right)}}}$\\\\\\ $\textbf{\underline{\underline{Given by Kaymant Sir...........}} }$\\\\\\