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Question Number 210933 by mnjuly1970 last updated on 22/Aug/24

$$$$$$$$$$$$$$$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{sin\left(25x\right)}{sinx}dx=?$$$$$$$$$$

Answered by Ghisom last updated on 22/Aug/24

$$\frac{\sin 25x}{\sin x}=\frac{\frac{{\mathrm{e}}^{25\mathrm{i}x}-{\mathrm{e}}^{-25\mathrm{i}x}}{2\mathrm{i}}}{\frac{{\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}}{2\mathrm{i}}}=$$$$=1+\underset{n=1}{\sum ^{12}}({\mathrm{e}}^{2nx\mathrm{i}}+{\mathrm{e}}^{-2nx\mathrm{i}})=$$$$=1+2\underset{n=1}{\sum ^{12}}\cos 2nx$$$$\Rightarrow$$$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sin 25x}{\sin x}dx={[x+\underset{n=1}{\sum ^{12}}\frac{\sin 2nx}{n}]}_0^{\pi /2}=$$$$=\frac{\pi }{2}$$

Commented by mnjuly1970 last updated on 23/Aug/24

$$thanksalotsir.excellent$$

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