$$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx \\|b|<1$$

I tried putting $-x$ using properties of definite integral, but that doesn't really help.

$$I=\int_{-\frac\pi2}^{\frac\pi2}\frac{-\ln(1-b\sin x)}{\sin x}dx$$

I don't think adding these 2 equations would yield anything. And I just cannot think of a good substitution.

Edit : Plugging b=1, it gives 4.9xxx . as $\pi ^2 \approx 10$, the value is $\pi \arcsin b$. Other values for b confirm this.

  • $\begingroup$ Maybe integration by parts? $\endgroup$
    – orion
    Mar 27, 2014 at 8:06
  • $\begingroup$ @orion I think that it will complicate it further. $\endgroup$
    – evil999man
    Mar 27, 2014 at 8:09
  • $\begingroup$ Even wolfram alpha isn't replying... $\endgroup$
    – evil999man
    Mar 27, 2014 at 8:30
  • 4
    $\begingroup$ $$I(b)=\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x}dx\iff I'(b)=\int_{-\frac\pi2}^{\frac\pi2}\frac{dx}{1+b\sin x}$$ $\endgroup$
    – Lucian
    Mar 27, 2014 at 9:16
  • $\begingroup$ I tried this but the new integral in b looked scary. $\endgroup$
    – lakshayg
    Mar 27, 2014 at 9:20

3 Answers 3


From the Taylor series expansion of $\ln(1+x)$ we know that $$\ln(1+b\sin x) = b\sin x - \frac{b^2\sin^2x}{2} +\frac{b^3\sin^3x}{3} - \cdots$$

Therefore $$\frac{\ln(1+b\sin x)}{\sin x} = b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots$$

Integrating both sides of the equation $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots \mathrm{d}x$$

Observe that $\int^{\pi/2}_{-\pi/2}\sin^nx\mathrm{d}x = 0$ if n is an odd number. The above equation is then equivalent to $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$

As $\sin^{2n}x$ is an even function therefore $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2 \int^{\pi/2}_0 b +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$

Using the fact that $\int_0^{\pi/2}\sin^nx\mathrm{d}x = \frac{(n-1)\cdot(n-3)\cdots3\cdot1}{n\cdot(n-2)\cdots4\cdot2}$ if n is even (See this page for the proof)

$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2\cdot\frac{\pi}{2}\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$ $$=\pi\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$ The term in the parenthesis is precisely the Taylor series expansion (Thanks to your edit. I was completely clueless at this point) for $\sin^{-1} b $. Therefore the integral is $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \pi\sin^{-1} b $$

  • $\begingroup$ Knowing the answer this looks like back working. $\endgroup$
    – evil999man
    Mar 27, 2014 at 9:04
  • $\begingroup$ Actually it in not. I checked it only after your edit that it was indeed the expansion for arcsin(b). Anyway, thanks a lot. $\endgroup$
    – lakshayg
    Mar 27, 2014 at 9:11
  • $\begingroup$ @awesome. I am curious to see if you get a better answer. $\endgroup$ Mar 27, 2014 at 9:12
  • $\begingroup$ We have to see how the hell someone made this question? $\endgroup$
    – evil999man
    Mar 27, 2014 at 9:13
  • $\begingroup$ @LakshayGarg. Thanks for this answer. I also started using Taylor expansions but they were based on $x$ and not $\sin(x)$ (you can see how stupid I can be or I am). Cheers. $\endgroup$ Mar 27, 2014 at 9:14

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\mbox{Let's}\ {\cal I}\pars{b} \equiv \int_{-\pi/2}^{\pi/2}{\ln\pars{1 + b\sin\pars{x}} \over \sin\pars{x}} \,\dd x:\ {\large ?}}$ with $\ds{b \in {\mathbb R}\,,\ \verts{b} < 1.\quad}$ $\ds{{\cal I}\pars{0} = 0}$

With Weierstrass Substitution $\ds{t \equiv \tan\pars{x \over 2}}$: \begin{align} {\cal I}'\pars{b}&=\int_{-\pi/2}^{\pi/2}{\dd x \over 1 + b\sin\pars{x}} =\int_{-1}^{1}{2\,\dd t/\pars{1 + t^{2}} \over 1 + b\bracks{2t/\pars{1 + t^{2}}}} =2\int_{-1}^{1}{\dd t \over t^{2} + 2bt + 1} \\[3mm]&=2\int_{-1}^{1}{\dd t \over \pars{t + b}^{2} + 1 - b^{2}} =2\int_{-1 + b}^{1 + b}{\dd t \over t^{2} + 1 - b^{2}} \\[3mm]&={2 \over \root{1 - b^{2}}} \int_{\pars{-1 + b}/\root{1 - b^{2}}}^{\pars{1 + b}/\root{1 - b^{2}}} {\dd t \over t^{2} + 1} \\[3mm]&={2 \over \root{1 - b^{2}}}\bracks{% \arctan\pars{1 + b \over \root{1 - b^{2}}} -\arctan\pars{-1 + b \over \root{1 - b^{2}}}} \\[3mm]&={2 \over \root{1 - b^{2}}}\bracks{% {\pi \over 2} - \arctan\pars{\root{1 - b^{2}} \over 1 + b} -\arctan\pars{-1 + b \over \root{1 - b^{2}}}} \\[3mm]&={2 \over \root{1 - b^{2}}}\braces{% {\pi \over 2} - \overbrace{% \bracks{\arctan\pars{1 - b \over \root{1 - b^{2}}} +\arctan\pars{-1 + b \over \root{1 - b^{2}}}}}^{\ds{=\ 0}}} \\[3mm]&={\pi \over \root{1 -b^{2}}} \end{align}

Then $$\color{#00f}{\large% \int_{-\pi/2}^{\pi/2}{\ln\pars{1 + b\sin\pars{x}} \over \sin\pars{x}}\,\dd x =\pi\,\arcsin\pars{b}} $$


I have been trying to compute the antiderivative; I tried several change of variable with absolutely nos success; as already mentioned, integration by parts makes the problem still more complex.

Using a CAS (brut force), the antiderivative which came out is just a nightmare for me $$-\text{Li}_2\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{b^2-1}-b}\right)-\text{Li}_2\left(-\frac{\tan \left(\frac{x}{2}\right)}{b+\sqrt{b^2-1}}\right)+\log \left(\tan \left(\frac{x}{2}\right)\right) \left(-\log \left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{b^2-1}+b}+1\right)-\log \left(\left(\sqrt{b^2-1}+b\right) \tan \left(\frac{x}{2}\right)+1\right)+\log (b \sin (x)+1)+\log \left(1-i \tan \left(\frac{x}{2}\right)\right)+\log \left(1+i \tan \left(\frac{x}{2}\right)\right)\right)+\frac{1}{2} \text{Li}_2\left(-\tan ^2\left(\frac{x}{2}\right)\right)$$ Computing the integral leads, after simplifications, to $$-\text{Li}_2\left(\frac{1}{\sqrt{b^2-1}-b}\right)-\frac{1}{2} \text{Li}_2\left(\frac{1}{\left(b+\sqrt{b^2-1}\right)^2}\right)+2 \text{Li}_2\left(\frac{1}{b+\sqrt{b^2-1}}\right)+\text{Li}_2\left(b+\sqrt{b^2-1}\right) $$ which is a real valued function of $b$. Trying to use the definition and properties of the dilogarithm took me to a dead end.

Thanks to your edit and clever remark, I verified that the result is exactly what you said, that is to say that $$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx = \pi \sin ^{-1}(b)$$

I checked that the Taylor series built at $b=0$ are identical.

Congratulations for having found the result. I am really sorry to have not been able to simplify these monsters.

Added later after Lucian's suggestion

After Lucian's so good suggestion, let us consider $$I'(b)=\int_{-\frac\pi2}^{\frac\pi2}\frac{dx}{1+b\sin x}$$ First of all, the antiderivative is given by $$\frac{2 \tan ^{-1}\left(\frac{b+\tan \left(\frac{x}{2}\right)}{\sqrt{1-b^2}}\right)}{\sqrt{1-b^2}}$$ Integrated between the given bounds $$I'(b)=\frac{2 \left(\tan ^{-1}\left(\sqrt{\frac{1-b}{1+b}}\right)+\tan ^{-1}\left(\sqrt{\frac{1+b}{1-b}}\right)\right)}{\sqrt{1-b^2}}=\frac{\pi }{\sqrt{1-b^2}}$$ and the final result for $I(b)$.

  • $\begingroup$ This should be not be that tough... $\endgroup$
    – evil999man
    Mar 27, 2014 at 9:02
  • $\begingroup$ @Awesome. I agree with you. Fortunately, you receive a nice answer from Lakshay Garg. $\endgroup$ Mar 27, 2014 at 9:10
  • $\begingroup$ @Claude. And my search for a better solution comes to an end. Well done. $\endgroup$
    – lakshayg
    Mar 27, 2014 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.