MHB Logarithmic Integral Calc: $$\int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)dx$$

[juantheron]

Calculation of $$\int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)dx$$

I Have tried like this way:: Let $$I = \int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)dx$$ Put $x=\sin \theta\;,$ Then $dx = \cos \theta d\theta$

and changing limits, we get $$I = \int_{0}^{\frac{\pi}{2}}\ln\left(1+\sin \theta \right)\cdot \cos \theta d\theta$$

Now How can I solve after that,

Thanks

 

[MarkFL]

You would actually have:

$$I=\int_{0}^{\frac{\pi}{2}}\ln\left(1+\cos(\theta)\right)\cdot \cos(\theta)\,d\theta$$

At this point I would try IBP, where:

$$u=\ln(1+\cos(\theta))\,\therefore\,du=-\frac{\sin(\theta)}{1+\cos(\theta)}\,d\theta$$

$$dv=\cos(\theta)\,d\theta\,\therefore\,v=\sin(\theta)$$

 

[Prove It]

Calculation of $$\int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)dx$$

I Have tried like this way:: Let $$I = \int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)dx$$ Put $x=\sin \theta\;,$ Then $dx = \cos \theta d\theta$

and changing limits, we get $$I = \int_{0}^{\frac{\pi}{2}}\ln\left(1+\sin \theta \right)\cdot \cos \theta d\theta$$

Now How can I solve after that,

Thanks

We should note that $\displaystyle \begin{align*} \textrm{arsech}\,{(x)} \equiv \ln{\left( \frac{1}{x} + \sqrt{ \frac{1}{x^2} - 1 } \right) } \textrm{ for } 0 < x \leq 1 \end{align*}$, so

$\displaystyle \begin{align*} \ln{ \left( 1 + \sqrt{ 1 - x^2 } \right) } &= \ln{ \left[ \frac{x\,\left( 1 + \sqrt{ 1 - x^2 } \right)}{x} \right] } \\ &= \ln{ \left[ x\,\left( \frac{1}{x} + \frac{\sqrt{1 - x^2}}{\sqrt{x^2}} \right) \right] } \\ &= \ln{ \left[ x\,\left( \frac{1}{x} + \sqrt{ \frac{1 - x^2}{x^2} } \right) \right] } \\ &= \ln{ \left[ x \, \left( \frac{1}{x} + \sqrt{ \frac{1}{x^2} - 1 } \right) \right] } \\ &= \ln{(x)} + \ln{ \left( \frac{1}{x} + \sqrt{ \frac{1}{x^2} - 1 } \right) } \\ &= \ln{(x)} + \textrm{arsech}\,(x) \end{align*}$

You should be able to look these integrals up in your tables :)

 

[MarkFL]

You would actually have:

$$I=\int_{0}^{\frac{\pi}{2}}\ln\left(1+\cos(\theta)\right)\cdot \cos(\theta)\,d\theta$$

At this point I would try IBP, where:

$$u=\ln(1+\cos(\theta))\,\therefore\,du=-\frac{\sin(\theta)}{1+\cos(\theta)}\,d\theta$$

$$dv=\cos(\theta)\,d\theta\,\therefore\,v=\sin(\theta)$$

Just to follow up after 24 hours...using my suggestion for continuing, we now have:

$$I=\left.\sin(\theta)\ln(1+\cos(\theta))\right|_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\sin^2(\theta)}{1+\cos(\theta)}\,d\theta$$

Using $$\frac{\sin^2(\theta)}{1+\cos(\theta)}=\frac{1-\cos^2(\theta)}{1+\cos(\theta)}=1-\cos(\theta)$$ we have:

$$I=0+\int_0^{\frac{\pi}{2}}1-\cos(\theta)\,d\theta$$

$$I=\left.\theta-\sin(\theta)\right|_0^{\frac{\pi}{2}}=\left(\frac{\pi}{2}-1\right)-(0-0)=\frac{\pi}{2}-1$$