Integration Questions

Question Number 138771 by qaz last updated on 18/Apr/21

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx}=? \\$$

Answered by Kamel last updated on 19/Apr/21

Answered by MJS_new last updated on 18/Apr/21

$$\int\mathrm{ln}\:\left(\mathrm{tan}^{\mathrm{4}} \:{x}\:−\mathrm{tan}^{\mathrm{2}} \:{x}\:+\mathrm{1}\right)\:{dx}= \\$$ $$\:\:\:\:\:\left[{t}={tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\$$ $$=\int\frac{\mathrm{ln}\:\left({t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\$$ $$\:\:\:\:\:\left[\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i},\:\beta=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i},\:\gamma=\overline {\alpha},\:\delta=\overline {\beta}\right] \\$$ $$=\int\frac{\mathrm{ln}\:\left({t}−\alpha\right)+\mathrm{ln}\:\left({t}−\beta\right)\:+\mathrm{ln}\:\left({t}−\gamma\right)\:+\mathrm{ln}\:\left({t}−\delta\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\$$ $$\mathrm{now}\:\mathrm{use} \\$$ $$\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=−\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}−\mathrm{i}}{dt}+\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}+\mathrm{i}} \\$$ $$\mathrm{and} \\$$ $$\int\frac{\mathrm{ln}\:\left({t}−{y}\right)}{{t}\pm{z}}{dt}=\mathrm{ln}\:\mid{x}\pm{z}\mid\:\mathrm{ln}\:\mid{y}\pm{z}\mid\:−\mathrm{Li}_{\mathrm{2}} \:\frac{{x}\pm{z}}{{y}\pm{z}} \\$$ $$... \\$$

Commented byqaz last updated on 18/Apr/21

$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx} \\$$ $$=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}\right)+{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$ $${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} −{ax}\right)+{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} +{ax}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$ $${I}\left(\mathrm{0}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{2}\pi{ln}\mathrm{2} \\$$ $${I}\left({a}\right)'=\int_{\mathrm{0}} ^{\infty} \left(\frac{−{x}}{\mathrm{1}+{x}^{\mathrm{2}} −{ax}}+\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{ax}}\right)\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\$$ $$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −{a}^{\mathrm{2}} {x}^{\mathrm{2}} }\right){dx} \\$$ $$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\left(\mathrm{2}−{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{1}}{dx}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\left(\mathrm{2}−{a}^{\mathrm{2}} \right)}{dx}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\left(\mathrm{4}−{a}^{\mathrm{2}} \right)}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right) \\$$ $${I}={I}\left(\sqrt{\mathrm{3}}\right)={I}\left(\mathrm{0}\right)+\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{2}}{{a}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right){da}=\mathrm{2}\pi{ln}\mathrm{2}−\pi{ln}\frac{\mathrm{4}}{\mathrm{3}}=\pi{ln}\mathrm{3} \\$$

Commented byMJS_new last updated on 18/Apr/21

$$\mathrm{great}!\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{think}\:\mathrm{of}\:\mathrm{this} \\$$

Commented byqaz last updated on 18/Apr/21

$${it}'{s}\:{true} \\$$

Commented byMJS_new last updated on 18/Apr/21

$$\mathrm{yes} \\$$