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Integrate using differentiation wrt parameter only.

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

We can express this as

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

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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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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