Question-200863

Question Number 200863 by sonukgindia last updated on 25/Nov/23

Answered by Mathspace last updated on 25/Nov/23

x^(2π) =t ⇒x=t^(1/(2π)) and I=∫_0 ^∞ (x^((π/5)−1) /(1+x^(2π) ))dx=(1/(2π))∫_0 ^∞ (((t^(1/(2π)) )^((π/5)−1) )/(1+t))t^((1/(2π))−1) =(1/(2π))∫_0 ^∞ (t^((1/(10))−(1/(2π))+(1/(2π))−1) /(1+t))dt =(1/(2π))∫_0 ^∞ (t^((1/(10))−1) /(1+t))dt=(1/(2π))×(π/(sin((π/(10))))) =(1/(2sin((π/(10)))))

$${x}^{\mathrm{2}\pi} ={t}\:\Rightarrow{x}={t}^{\frac{\mathrm{1}}{\mathrm{2}\pi}} \:{and} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{\pi}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}\pi} }{dx}=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\infty} \frac{\left({t}^{\frac{\mathrm{1}}{\mathrm{2}\pi}} \right)^{\frac{\pi}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{t}}{t}^{\frac{\mathrm{1}}{\mathrm{2}\pi}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\frac{\mathrm{1}}{\mathrm{2}\pi}+\frac{\mathrm{1}}{\mathrm{2}\pi}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\mathrm{1}}{\mathrm{2}\pi}×\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{10}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{10}}\right)} \\ $$

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Question-200863

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