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Question Number 799 by 123456 last updated on 15/Mar/15
∫_0 ^π sin x ln (1−cos x)dx converge?
$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:{x}\:\mathrm{ln}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right){dx}\:{converge}? \\ $$
Answered by prakash jain last updated on 15/Mar/15
1−cos x=t  x=0⇒t=0  x=π, t=2  sin x dx=dt  ∫ln t dt=t(ln t−1)  lim_(t→0) (2ln 2−2−tln t+t)  =ln 4−2−lim_(t→0) tln t  =ln 4−2−0  =ln 4−2
$$\mathrm{1}−\mathrm{cos}\:{x}={t} \\ $$$${x}=\mathrm{0}\Rightarrow{t}=\mathrm{0} \\ $$$${x}=\pi,\:{t}=\mathrm{2} \\ $$$$\mathrm{sin}\:{x}\:{dx}={dt} \\ $$$$\int\mathrm{ln}\:{t}\:{dt}={t}\left(\mathrm{ln}\:{t}−\mathrm{1}\right) \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{2ln}\:\mathrm{2}−\mathrm{2}−{t}\mathrm{ln}\:{t}+{t}\right) \\ $$$$=\mathrm{ln}\:\mathrm{4}−\mathrm{2}−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}{t}\mathrm{ln}\:{t} \\ $$$$=\mathrm{ln}\:\mathrm{4}−\mathrm{2}−\mathrm{0} \\ $$$$=\mathrm{ln}\:\mathrm{4}−\mathrm{2} \\ $$

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