Integration Questions

Question Number 187135 by cortano12 last updated on 14/Feb/23

$$\:\underset{\:\:\mathrm{0}} {\overset{\:\:\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx}\:=? \\$$

Answered by horsebrand11 last updated on 14/Feb/23

$$\:{I}=\underset{\:\mathrm{0}} {\overset{\:\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx}\: \\$$$$\:{Let}\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}} \\$$$$\:\begin{cases}{\mathrm{tan}\:{x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}}\\{{dx}=−\left(\mathrm{1}+\mathrm{sin}\:{x}\right){dt}}\end{cases} \\$$$$\:\lambda\:=\int\:\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right)^{\mathrm{2}} {dt}=\int\left(\frac{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }\right){dt} \\$$$$\:\lambda=\:\int\:\left(\frac{\mathrm{1}}{\mathrm{4}}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}{t}^{−\mathrm{2}} \right){dt} \\$$$$\:\lambda=\frac{\mathrm{1}}{\mathrm{12}}{t}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\mathrm{1}}{\mathrm{4}{t}}\:+{c}\: \\$$$${I}=\:\left[\:\frac{\mathrm{1}}{\mathrm{12}}{t}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\mathrm{1}}{\mathrm{4}{t}}\:\right]_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\mathrm{1}} \\$$$${I}=\left(\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\left(\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{12}}−\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}\right) \\$$$${I}=\:\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{3}} \\$$