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$$\int^{\pi/3}_{\pi/4} \frac{\tan^2x}{x-\tan x} dx $$

this is that I have tried

$$\int^{\pi/3}_{\pi/4} \frac{\frac{\sin^2x}{\cos^2 x}}{x-\frac{\sin x}{\cos x}} dx $$

$$\int^{\pi/3}_{\pi/4} \frac{\sin^2(x)}{x \cos^2(x)-\sin(x)\cos(x)} dx $$

but I an not making anymore progress

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Hint: For the integral

$$\int \dfrac{\tan^2(x)}{x-\tan(x)}\, dx$$

try the substitution $u = x-\tan(x)$, $du = 1-\frac{1}{\cos(x)^2} \, dx = -\tan^2(x)\, dx$

As requested here is the elaboration:

With the substitution $u = x-\tan(x)$ we get \begin{align*}du &= \dfrac{d}{dx}\left(x-\frac{\sin(x)}{\cos(x)}\right)dx=1-\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}dx=1-\frac{1}{\cos^2(x)}dx\\&=\frac{\cos^2(x)-1}{\cos^2(x)}dx=\frac{-\sin^2(x)}{\cos^2(x)}dx=-\tan^2(x)\, dx\end{align*}

So your integral becomes

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\dfrac{\tan^2(x)}{x-\tan(x)}\, dx=-\int_{\frac{\pi}{4}-1}^{\frac{\pi}{3}-\sqrt{3}}\frac{1}{u}\,du=\ln\left(\frac{\pi}{3}-\sqrt{3}\right)-\ln\left(\frac{\pi}{4}-1\right)$$

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  • $\begingroup$ will try that thanx $\endgroup$
    – Gobabis
    Sep 18 '14 at 11:26
  • $\begingroup$ @Gobabis ask if you have troubles :) $\endgroup$
    – Bman72
    Sep 18 '14 at 11:27
  • $\begingroup$ @Ale. Could you elaborate, please ? $\endgroup$ Sep 18 '14 at 11:32
  • $\begingroup$ @ClaudeLeibovici Yes, I've added some steps, do you see something wrong? If yes let me know please :) $\endgroup$
    – Bman72
    Sep 18 '14 at 11:50
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    $\begingroup$ @Ale. Thank you very much. It is now a very beautiful solution. Congratulations. Cheers :-) $\endgroup$ Sep 18 '14 at 16:16

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