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\begin{align}\int_{{\pi}\over5}^{{3\pi}\over10}\frac{x}{\sin2x}\,dx\end{align} This integral came up while learning integration using Leibnitz rule. What has been tried is taking the integral as: \begin{align}I(a)=\frac{1}{2}\int_{\pi\over5}^{3\pi\over10}\frac{\arctan (a \tan x)}{\sin x\cos x}\,dx\end{align} which gives $I'(a)$ as: \begin{align}I'(a)=\frac{1}{2}\int_{2\pi\over5}^{3\pi\over5}\frac{dx}{\cos^2x(1+a^2\tan^2x)}=\frac{1}{2a}\arctan{\frac{a\tan \frac{3\pi}{10}-a\tan{\pi\over5}}{1+a^2\tan{\pi\over5}\cot{\pi\over5}}}\end{align} \begin{align}=\frac{1}{2a}\arctan{\frac{a(\tan \frac{3\pi}{10}-\tan{\pi\over5})}{1+a^2}}\end{align} Proceeding after this seems uncertain.

Shimura Variety
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    This Wolfram Alpha Output suggests that something is off. Special functions? I am not sure if WA is right on this one, but it's wrong very rarely so I would request you to check the question once again , and see if it's the right one. – Sarvesh Ravichandran Iyer Jun 09 '22 at 13:33
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    I have checked; maybe there was a typo from the writer's side,because it is intended for highschool students preparing for JEE, but wolfram alpha gives an antiderivative nonetheless, so maybe it will be fine to just wait for answers and try it in the meantime. – Shimura Variety Jun 09 '22 at 13:55
  • @DevanshBhardwaj can you provide some additional context? Say, the previous and following two lines wrt the integral(of it is mentioned in the text), or the problem you need the integral to solve? – insipidintegrator Jun 09 '22 at 14:06
  • The problem is without words @insipidintegrator. It is simply the integration problem. There is a kind of answer given in terms of a function with two values (e.g., as of b and c in $\frac{\ln b}{\tan c}$) that are numbers we have to find; but that would only bias the view of the answerer, thus not making it possible to see how we can actually approach it independently, so I am leaving it out. However, it is made by high school students only, and we are meant to integrate it fully; the solution is given in that form because of exam pattern of numericals (e.g, they would ask 6bc.) – Shimura Variety Jun 09 '22 at 14:39
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    While the answer given is a better approach, the next step in your method would be to use the identity $\arctan(kx)-\arctan(x/k) = \arctan((k-1/k) x/(1+x^2))$. – eyeballfrog Jun 09 '22 at 15:37

1 Answers1

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A simpler approach using the following property of definite integral:

$$\int_{a}^ {b} f(x) dx=\int_{a} ^{b} f(a+b-x)dx$$

\begin{align} I&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{x}{\sin{2x}}\:dx\\ &=\int_{\pi \over 5}^{3 \pi \over 10}\frac{{\pi \over 5}+{3 \pi \over 10}-x}{\sin\left({2\left({\pi \over 5}+{3 \pi \over 10}-x\right)}\right)}\: dx\\ &=\int_{\pi \over 5}^{3 \pi \over 10}\frac{\frac{\pi}{2}-x}{\sin{2\left(\frac{\pi}{2}-x\right)}}\:dx\\ &=\int_{\pi \over 5}^{3 \pi \over 10}\frac{\frac{\pi}{2}}{\sin{2x}}\:dx -\int_{\pi \over 5}^{3 \pi \over 10}\frac{x}{\sin{2x}}\:dx\\ &=\frac{\pi}{2}\int_{\pi \over 5}^{3 \pi \over 10}\frac{dx}{\sin{2x}} -I\\ \implies 2I &=\frac{\pi}{2}\int_{\pi \over 5}^{3 \pi \over 10} \csc{2x}\: dx\\ &=\frac{\pi}{2}\frac{1}{2} \left[\ln\left|\tan\left(\frac{2x}2\right)\right|\right]_{\pi \over 5}^{3 \pi \over 10}\\ &=\frac{\pi}{4} \left(\ln\left|\tan\left(\frac{3\pi}{10}\right)\right|-\ln\left|\tan\left(\frac{\pi}{5}\right)\right|\right)\\ \implies I&=\frac{\pi}{8} \ln\left|\frac{\tan\left(\frac{3\pi}{10}\right)}{\tan\left(\frac{\pi}{5}\right)}\right|\\ &\approx 0.250901 \end{align}

Aman Kushwaha
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