Integration Questions

Question Number 200685 by Spillover last updated on 21/Nov/23

$$\:\:\:\:\:\: \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\:\left(\mathrm{1}+\mathrm{tan}{x}\right){dx}\: \\$$$$\\$$

Answered by som(math1967) last updated on 22/Nov/23

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