Integration Questions

Question Number 122963 by Lordose last updated on 21/Nov/20

$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\$$

Answered by Dwaipayan Shikari last updated on 21/Nov/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{log}\left(\mathrm{1}+{tan}\theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}.{sec}^{\mathrm{2}} \theta{d}\theta \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left(\sqrt{\mathrm{2}}\left({cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right)−{log}\left({cosx}\right){dx}\right. \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right)−{log}\left({cosx}\right) \\$$ $$=\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)+{I} \\$$ $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right)−{log}\left({cosx}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cosx}\right)−{log}\left({cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right) \\$$ $${I}=\mathrm{0} \\$$ $${So}\:{answer}\:{is}\: \\$$ $$\Rightarrow\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right) \\$$

Answered by mnjuly1970 last updated on 21/Nov/20

$$\:\:{solution}: \\$$ $$\:\:{x}={tan}\left({y}\right)\Rightarrow\:{dx}=\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({y}\right)\right){dy} \\$$ $${I}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left({y}\right)\right){dy} \\$$ $$=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left[\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{y}\right)\right]{dy} \\$$ $$=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}−{tan}\left({y}\right)}{\mathrm{1}+{tan}\left({y}\right)}\right){dy} \\$$ $$=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \left\{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{1}+{tan}\left({y}\right)\right)\right\}{dy} \\$$ $$=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−{I} \\$$ $$\mathrm{2}{I}=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:\Rightarrow{I}=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\:\checkmark \\$$