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Question Number 133734 by Abdoulaye last updated on 23/Feb/21

$$howdowecaculate\int ln\left(tanx\right)dx?$$

Answered by mathmax by abdo last updated on 24/Feb/21

$$\mathrm{f}\left(\mathrm{x}\right)={\int }_{\frac{\pi }{4}}^{\mathrm{x}}\ln \left(\mathrm{tanx}\right)\mathrm{dx}\Rightarrow \mathrm{f}\left(\mathrm{x}\right){=}_{\mathrm{tanx}=\mathrm{t}}{\int }_1^{\mathrm{tanx}}\frac{\ln \left(\mathrm{t}\right)}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$={\left[\arctan \left(\mathrm{t}\right)\ln \left(\mathrm{t}\right)\right]}_1^{\mathrm{tanx}}-{\int }_1^{\mathrm{tanx}}\frac{\mathrm{arctant}}{\mathrm{t}}\mathrm{dt}$$$$=\mathrm{xln}\left(\mathrm{tanx}\right)-{\int }_1^{\mathrm{tanx}}\frac{\arctan \left(\mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{have}{\left(\mathrm{arctant}\right)}^{\left(1\right)}=\frac{1}{1+{\mathrm{t}}^2}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{t}}^{2\mathrm{n}}\Rightarrow \arctan \left(\mathrm{t}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{t}}^{2\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)\Rightarrow$$$$\frac{\mathrm{arctant}}{\mathrm{t}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{t}}^{2\mathrm{n}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{xln}\left(\mathrm{x}\right)-{\int }_1^{\mathrm{tanx}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{t}}^{2\mathrm{n}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\left[\frac{1}{2\mathrm{n}+1}{\mathrm{t}}^{2\mathrm{n}+1}\right]}_1^{\mathrm{tanx}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^2}({\tan }^{2\mathrm{n}+1}\mathrm{x}-1)$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^2}{\tan }^{2\mathrm{n}+1}\mathrm{x}-\mathrm{K}\left(\mathrm{katalan}\mathrm{constant}\right)....\mathrm{be}\mathrm{continued}...$$

Commented by Abdoulaye last updated on 24/Feb/21

$$thank$$$$pleasewhatkatalancanstant?$$

Commented by Dwaipayan Shikari last updated on 24/Feb/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^2}=1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+....=G\left({Catalan}^{\prime }sconstant\right)$$

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