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In my previous Question [1], @Quanto has defined

$$J(a) = \int_0^{\frac\pi2}\ln(1+\sin a\sin x)\,dx$$ and stated $$J'(a) =\int_0^{\frac\pi2}\frac{\cos a\sin x}{1+\sin a\sin x}\,dx =a\csc a-\frac\pi2\tan \frac a2 $$

But, I was wondering if this is valid $\forall a\in \Bbb R-{{n\pi}}$.

Using Desmos, it seems that this is valid for $a\in \Big(-\frac{3\pi}{2},\frac{\pi}{2}\Big)$. Is there any explanation can we offer?

On desmos, I run $a$ from $-1000$ to $1000$ and I see that the the Integral $\Big(J'(a)\Big)$ achieves value only from $-\pi$ to $\pi$.

Michael Hardy
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User 1207
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2 Answers2

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Note

\begin{align} I=\int_0^{\frac\pi2}\frac{\cos a\sin x}{1+\sin a\sin x}dx = \cot a\left( \frac\pi2 - \int_0^{\frac\pi2}\frac{1}{1+\sin a\sin x}dx\right)\tag1\\ \end{align} With $\cos x= \frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}$ \begin{align}& \int_0^{\frac\pi2}\frac{1}{1+\sin a\sin x}dx =\int_0^{\frac\pi2}\frac{1}{1+\sin a\cos x}dx \\ = &\>2\int_0^{\frac\pi2}\frac{d(\tan\frac x2)}{(1-\sin a)\tan^2\frac x2+(1+\sin a)}\\ = &\>2\sec a \tan^{-1}\left(\tan(\frac\pi4 -\frac a2)\right) =\>\sec a \>\left(\frac\pi2 -a \right)\tag2\\ \end{align} where the last step requires $\frac\pi4 -\frac a2\in (-\frac\pi2,\frac\pi2)$, or $a\in ( - \frac\pi2, \frac{3\pi}2)$. Substitute (2) into (1) to arrive at $$I=a\csc a-\frac\pi2\tan \frac a2$$

Quanto
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Using the tangent half-angle substitution, $$\int\frac{\cos (a)\sin (x)}{1+\sin (a)\sin (x)}dx=x \cot (a)+2 \csc (a) \tan ^{-1}\Bigg[\cos \left(\frac{a-x}{2}\right) \csc \left(\frac{a+x}{2}\right)\Bigg]$$ $$\int_0^{\frac\pi2}\frac{\cos (a)\sin (x)}{1+\sin (a)\sin (x)}dx=\frac{1}{2} \pi \cot \left(\frac{a}{2}\right)-2 \csc (a) \tan ^{-1}\Big[\cot \left(\frac{a}{2}\right)\Big]$$

Being lazy, I asked a CAS for restrictions and it gave $$\Re\left(\cos ^{-1}(\csc (a))\right)\geq \pi \lor \Re\left(\sin ^{-1}(\csc (a))\right)>0\lor \sin ^{-1}(\csc (a))\notin \mathbb{R}$$

Claude Leibovici
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    Can you please explain the meaning of last para(If it is easy to understand)? Being self-taught student I didn't understand them. – User 1207 Jul 01 '21 at 06:03
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    @User1207: If you're asking about the symbols, $\mathfrak R(z)$ denotes the real part of a complex number, which you might have seen as $\text{Re}(z)$. And, the symbol $\vee$ denotes "or". For example, $p\vee q$ means either p is true or q is true (or both). – ultralegend5385 Jul 01 '21 at 06:17
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    @ultralegend5385 Thanks – User 1207 Jul 01 '21 at 06:20