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\displaystyle \begin{aligned}\int_0^{\frac{1}{2}} \frac{\ln (1-x)}{x}\,dx\overset{x\to \frac{x}{2}}=\int_0^1 \frac{\ln (1-\frac{x}{2})}{x}\,dx \overset{ \text{IBP}}=\int_0^1 \frac{\ln x}{2-x}\,dt \overset{x\to 1-x}=\int_0^1 \frac{\ln (1-x)}{1+x}\,dx
\begin{aligned} \beta_n = \displaystyle \left[\frac{\cos \left(\frac{t}{2}(2n+1)\right) \left(\csc\frac{t}{2}-\frac{2}{t}\right)}{2n+1}\right]_0^{x} +\frac{1}{2n+1}\int_0^x\cos \left( \frac{t}{2}(2n+1)\right)\left( \cot \frac{t}{2} \csc\frac{t}{2}-\frac{4}{t^2}\right)\,dt\to 0
\displaystyle\begin{aligned} \int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \sum_{v\geq 1}\frac{\sin vt_{2k}}{v}\,dt_{2k}\cdots dt_1=\frac{1}{2}\int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \pi -t_{2k}\, dt_{2k}\cdots\, dt_1
\displaystyle\begin{aligned} (-1)^k\sum_{v\geq 1} \frac{\sin vx}{v^{2k+1}}+\sum_{i=1}^{k} \left( \frac{(-1)^{k+i+1}x^{2(k-i)+1}}{(2(k-i)+1)!}\sum_{v\geq 1}\frac{1}{v^{2i}} \right)= \frac{x^{2k}}{2}\left(\frac{\pi}{(2k)!}-\frac{x}{(2k+1)!}\right)
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Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71