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Question Number 98463 by pranesh last updated on 14/Jun/20

Answered by maths mind last updated on 15/Jun/20

$$\underset{k=1}{\sum ^{99}}\frac{x^k}{k}=\int \frac{1-x^{99}}{1-x}dx$$$$(cot\left(x\right)+......+\frac{{cot}^{99}\left(x\right)}{99})+\int (1+cot\left(x\right))(1+{cot}^{99}\left(x\right))dx$$$$\sum _{n⩾1}\frac{{cot}^n\left(x\right)}{n}=\sum _{n⩾1}\frac{t^n}{n}=\int \frac{1-t^{99}}{1-t}dt$$$$\mathrm{\Sigma }\frac{{cot}^k\left(x\right)}{k}dx=\int \frac{1-{cot}^{99}\left(x\right)}{1-cot\left(x\right)}.(-1-{cot}^2\left(x\right))dx$$$$weget$$$$\int \frac{1-{cot}^{99}\left(x\right)}{1-cot\left(x\right)}(-1-{cot}^2\left(x\right))dx+\int (1+cot\left(x\right))(1+{cot}^{99}\left(x\right))dx$$$$=\int \frac{({cot}^{99}\left(x\right)-1)(1+{cot}^2\left(x\right))+(1-{cot}^2\left(x\right))(1+{cot}^{99}\left(x\right))}{1-cot\left(x\right)}dx$$$$=\int \frac{2{cot}^{99}\left(x\right)-2{cot}^2\left(x\right)}{1-cot\left(x\right)}dx$$$$=-2\int {cot}^2\left(x\right)\frac{1-{cot}^{97}\left(x\right)}{1-cot\left(x\right)}dx$$$$=-2\int {cot}^2\left(x\right).\underset{k=0}{\sum ^{96}}{cot}^k\left(x\right)dx$$$$=-2\int [{cot}^2\left(x\right)+{cot}^3\left(x\right)+.......+{cot}^{98}\left(x\right)]dx$$$$so$$$$\lambda =2$$$$$$

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