# Evaluate $\int_{0}^{1} \frac{\ln x}{(1+8x^2)\sqrt{1-x^2}} dx$

Stuck in by parts ( or any other method is appreciated ): $$\int_{0}^{1}\frac{\ln\left(x\right)} {\left(1 + 8x^{2}\right)\sqrt{1 - x^{2}}}\,{\rm d}x$$

• To evaluate this integral, I have tried many different ideas but none seems to get me to any result.
• I have tried $$x = \sin\left(t\right):\ \mbox{integral of}\quad \frac{\ln\left(\sin\left(x\right)\right)}{1 + 8 \sin^{2}\left(x\right)},$$ then by parts taking $$\ln\left(\sin\left(x\right)\right)$$ as first function, but I am getting stuck in integration by parts of $$\cot\left(x\right)\arctan\left(3\tan\left(x\right)\right)$$ Please help.
• Integral of (ln sinx)/(1+ 8 sin²x), then taking ln sinx as first function, but stuck in integral of cotx arctan(3tanx) Commented Jul 9 at 6:35
• Alternatively took lnx as first function and remaining as second function, still getting an inverse trigo function divided by x. Commented Jul 9 at 6:36
• I doubt if it's elementary: wolframalpha.com/… Commented Jul 9 at 6:38
• So should I use Simpson's rule and just approximate it? Commented Jul 9 at 6:39
• I might do that, or rewrite the integral before approximating it. Commented Jul 9 at 6:44

$$I=\int_0^1\frac{log(x)}{(1 + 8 x^2) \sqrt{1 - x^2}}\,dx$$

Put $$x=\cos\theta$$ $$I=\int_0^\frac\pi2 \frac{ln(\cos\theta)}{1+8\cos^2\theta}\,d\theta$$ Put $$t=\tan\theta$$

$$I=\int_0^{\infty}\frac{ln\left(\frac{1}{\sqrt{1+t^2}}\right)}{1+\left(\frac{8}{{1+t^2}}\right)}\frac{1}{1+t^2}\,dt=\frac{-1}{2}\int_0^{\infty}\frac{ln(1+t^2)}{9+t^2}\,dt$$

$$\boxed{\int_{0}^{\infty} \frac{\ln(a^{2}+x^{2})}{b^{2}+x^{2}} \, dx = \frac{\pi}{b} \, \ln(a+b)}$$

The above result is from here

Set $$a=1,b=3$$ in the above formula;

$$I=-\frac{\pi}{3} \, \ln(2)\approx -0.725$$

Which also agrees with this

Starting with OP’s substitution, $$x=\sin \theta$$, then $$I=\int_0^{\frac{\pi}{2}} \frac{\ln (\sin \theta)}{1+8 \sin ^2 \theta} d \theta$$ Let $$t=\cot \theta$$, we have \begin{aligned} I & =\int_{\infty}^0 \frac{\ln \left(\frac{1}{\sqrt{1+t^2}}\right)}{1+\frac{t}{1+t^2}} \cdot \frac{-d t}{1+t^2} \\ & =-\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{9+t^2} d t \\ & =-\frac{1}{4} \int_{-\infty}^{\infty} \frac{\ln \left(1+t^2\right)}{9+t^2} d t \\ & =-\frac{1}{4}\cdot 2 \Re \int_{-\infty}^{\infty} \frac{\ln (t+i)}{9+t^2} d t \\ & =-\frac{1}{4} \cdot 2 \Re\left[2 \pi i \cdot \lim_{t\rightarrow 3i}(t-3 i) \frac{\ln(t+i)}{t^2+9 }\right] \\ & =-\frac{1}{2} \Re\left[2 \pi i \cdot \frac{\ln (4 i)}{6 i}\right] \\ & =-\frac{\pi}{3} \ln 2 \end{aligned}

Let $$t=\frac{\sqrt {1-x^2}}{x}$$, then $$x^2=\frac{1}{1+t^2}$$ and \begin{aligned}\int_0^1 \frac{\ln x}{\left(1+8 x^2\right) \sqrt{1-x^2}} d x = & \int_{\infty}^0 \frac{\ln \left(\frac{1}{\sqrt{1+t^2}}\right)}{\left(1+\frac{8}{1+t^2}\right)\left(\frac{t}{\sqrt{1+t^2}}\right)} \cdot\left(-\frac{t}{\left(1+t^2\right)^{\frac 32}}\right) dt\\ = & -\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{9+t^2} d t \\ = & -\frac{1}{2} \cdot \frac{2 \pi\ln 2}{3} \quad \text { (as above) } \\ = & -\frac{\pi\ln 2}{3} \end{aligned}