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Question Number 151246 by mathmax by abdo last updated on 19/Aug/21

find I=∫_0 ^(π/4) ln(cosx)dx and J=∫_0 ^(π/4) ln(sinx)dx

$$\mathrm{find}\:\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx}\:\mathrm{and}\:\mathrm{J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx} \\ $$

Answered by qaz last updated on 19/Aug/21

∫_0 ^(π/4) lnsin xdx =(1/2)∫_0 ^(π/2) lnsin (x/2)dx =(1/4)∫_0 ^(π/2) ln((1−cos x)/2)dx =(1/4)∫_0 ^(π/2) ln(sin xtan (x/2))dx−(1/4)∫_0 ^(π/2) ln2dx =−(π/8)ln2+(1/4)∫_0 ^(π/2) lntan (x/2)dx−(π/8)ln2 =−(π/4)ln2+(1/2)∫_0 ^(π/4) lntan xdx =−(π/4)ln2+(1/2)∫_0 ^1 ((lnx)/(1+x^2 ))dx =−(π/4)ln2+(1/2)Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) lnxdx =−(π/4)ln2+(1/2)Σ_(n=0) ^∞ (((−1)^(n+1) )/((2n+1)^2 )) =−(π/4)ln2−(1/2)G −−−−−−−−−−−−−−− ∫_0 ^(π/4) lncos xdx+∫_0 ^(π/4) lnsin xdx =∫_0 ^(π/4) ln((1/2)sin 2x)dx =∫_0 ^(π/4) ln(1/2)dx+(1/2)∫_0 ^(π/2) lnsin xdx =−(π/2)ln2 ⇒∫_0 ^(π/4) lncos xdx=−(π/2)ln2−(−(π/4)ln2−(1/2)G)=(1/2)G−(π/4)ln2

$$\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{lnsin}\:\mathrm{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{lnsin}\:\frac{\mathrm{x}}{\mathrm{2}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{2}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\left(\mathrm{sin}\:\mathrm{xtan}\:\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln2dx} \\ $$$$=−\frac{\pi}{\mathrm{8}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{lntan}\:\frac{\mathrm{x}}{\mathrm{2}}\mathrm{dx}−\frac{\pi}{\mathrm{8}}\mathrm{ln2} \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{lntan}\:\mathrm{xdx} \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2n}} \mathrm{lnxdx} \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{G} \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{lncos}\:\mathrm{xdx}+\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{lnsin}\:\mathrm{xdx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2x}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{ln}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{lnsin}\:\mathrm{xdx} \\ $$$$=−\frac{\pi}{\mathrm{2}}\mathrm{ln2} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{lncos}\:\mathrm{xdx}=−\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\left(−\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{G}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{G}−\frac{\pi}{\mathrm{4}}\mathrm{ln2} \\ $$

Commented by peter frank last updated on 19/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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