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Integrate using differentiation wrt parameter only.

$$\int_0^{\pi/2} x\cot(x)dx$$

We can express this as

$$\int_0^{\pi/2} x\cdot\frac{\cos(x)}{\sin(x)}dx$$

Notice we can write $u=\sin(x)$ to start but I am not sure if that will do us any good.

If we use $\xi$ as a parameter, the answer is of the form.$$ \lim_{\xi \to 1}I(\xi)=\lim_{\xi \to 1}\frac{\pi}{2}\ln(\xi+1)$$

NOTE:Only use differentiation with respect to parameter.

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  • $\begingroup$ What is differentiation w.r.t. parameter? In any case, $\int x \cot x dx = x\int \cot x dx - \int (\dfrac{dx}{dx}(\int \cot x dx)) = x (\ln(\sin x)) - \int \ln(\sin x) dx + c$. The second integral with limits $0$ to $\pi/2$ is $-\ln 2 \dfrac{\pi}{2}$ and take limit as $x \to 0$ in first integral. $\endgroup$ – taninamdar Feb 27 '14 at 4:33
  • $\begingroup$ Where does appear the parameter you want to differentiate with respect to ? Where does appear $\xi$ ? Please clarify. $\endgroup$ – Claude Leibovici Feb 27 '14 at 4:47
  • $\begingroup$ @ClaudeLeibovici The point of the problem is to find a function you can introduce a parameter to. Thus I don't know where $\xi$ appears, that is the point of the problem lol $\endgroup$ – Jeff Faraci Feb 27 '14 at 16:00
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found the way to go...We can want to solve

$$ I=\int_0^{\pi/2}x\cot(x) dx, $$ so we introduce a parameter $\xi$ by writing

$$ I(\xi)=\int_0^{\pi/2} \frac{\tan^{-1}(\xi\tan(x))}{\tan(x)} dx $$ and in the limit $\xi \to 1$ we recover I. Taking a derivative we obtain $$ I'(\xi)=\frac{d}{d\xi}\int_0^{\pi/2} \frac{\tan^{-1}(\xi\tan(x))}{\tan(x)} dx=\int_0^{\pi/2}\frac{\partial}{\partial \xi} \bigg(\frac{\tan^{-1}(\xi\tan(x))}{\tan(x)} \bigg)dx $$ Now we take the derivative to obtain $$ I'(\xi)=\int_0^{\pi/2} \frac{dx}{\big(\xi\tan(x)\big)^2+1} =\frac{\pi}{2(\xi+1)}. $$ We now integrate our result wrt $\xi$ and realizing the constant of integration is zero, we obtain $$ I(\xi)=\frac{\pi}{2}\ln(\xi+1). $$ Taking the limit as $\xi \to 1$ we obtain $$ \lim_{\xi \to 1} I(\xi)=\lim_{\xi \to 1} \frac{\pi}{2}\ln(\xi+1)=\frac{\pi \ln(2)}{2}. $$ Thus we have shown that $$ {\boxed{I=\frac{\pi\ln(2)}{2}}} $$

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