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Question Number 191049 by Mingma last updated on 16/Apr/23
Commented by Mingma last updated on 17/Apr/23
Nice solution, sir!
Answered by aleks041103 last updated on 17/Apr/23
cos(sin(x))=Re(exp(isin(x)))⇒cos(sin(x))exp(cos(x))==Re(exp(cos(x)+isin(x)))==Re(eeix)⇒∫02πcos(sin(x))exp(cos(x))dx==Re(∫02πeeixdx)∫02πeeixdx=?z=eix⇒dz=izdx⇒dx=−idzz⇒∫02πeeixdx=−i∮ΓezzdzwhereΓ:∣z∣=1Byresiduetheorem:∮Γezzdz=2πiRes(z=0)Res(z=0)=limz→0ezz(z−0)=1⇒⇒∫02πeeixdx=−i(2πi)=2π⇒∫02πcos(sin(x))exp(cos(x))dx=2π
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