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Question-212314




Question Number 212314 by Spillover last updated on 09/Oct/24
Answered by mehdee7396 last updated on 09/Oct/24
f(x)=ln(((2−sinx)/(2+sinx)))  f(−x)=ln(((2+sinx)/(2−sinx)))=−f(x)  ⇒;f; is  ;odd⇒∫_(−(π/2)) ^(π/2) f(x)dx=0 ✓
$${f}\left({x}\right)={ln}\left(\frac{\mathrm{2}−{sinx}}{\mathrm{2}+{sinx}}\right) \\ $$$${f}\left(−{x}\right)={ln}\left(\frac{\mathrm{2}+{sinx}}{\mathrm{2}−{sinx}}\right)=−{f}\left({x}\right) \\ $$$$\Rightarrow;{f};\:{is}\:\:;{odd}\Rightarrow\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right){dx}=\mathrm{0}\:\checkmark \\ $$$$ \\ $$
Answered by Ghisom last updated on 09/Oct/24
f(x)=ln ((2−sin x)/(2+sin x))  f(−x)=ln ((2+sin x)/(2−sin x)) =−ln ((2−sin x)/(2+sin x)) =−f(x)  ⇒ ∫_(−a) ^a f(x)dx=0
$${f}\left({x}\right)=\mathrm{ln}\:\frac{\mathrm{2}−\mathrm{sin}\:{x}}{\mathrm{2}+\mathrm{sin}\:{x}} \\ $$$${f}\left(−{x}\right)=\mathrm{ln}\:\frac{\mathrm{2}+\mathrm{sin}\:{x}}{\mathrm{2}−\mathrm{sin}\:{x}}\:=−\mathrm{ln}\:\frac{\mathrm{2}−\mathrm{sin}\:{x}}{\mathrm{2}+\mathrm{sin}\:{x}}\:=−{f}\left({x}\right) \\ $$$$\Rightarrow\:\underset{−{a}} {\overset{{a}} {\int}}{f}\left({x}\right){dx}=\mathrm{0} \\ $$

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