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Question Number 31501 by abdo imad last updated on 09/Mar/18

$$find{\int }_0^{\frac{\pi }{4}}ln(1+2tanx)dx.$$

Commented by abdo imad last updated on 12/Mar/18

$$letintroducef\left(t\right)={\int }_0^{\frac{\pi }{4}}ln(1+ttanx)dx\Rightarrow$$$$f^{{}^{\prime }}\left(t\right)={\int }_0^{\frac{\pi }{4}}\frac{tanx}{1+ttanx}dx=\frac{1}{t}{\int }_0^{\frac{\pi }{4}}\frac{ttanx+1-1}{1+ttanx}dx$$$$=\frac{\pi }{4t}-\frac{1}{t}{\int }_0^{\frac{\pi }{4}}\frac{dx}{1+ttanx}butwehave$$$${\int }_0^{\frac{\pi }{4}}\frac{dx}{1+ttanx}={\int }_0^{\frac{\pi }{4}}\frac{cosx}{cosx+tsinx}dxletusethech.$$$$tan\left(\frac{x}{2}\right)=u\Rightarrow {\int }_0^{\frac{\pi }{4}}\frac{cosx}{cosx+tsinx}dx$$$$={\int }_0^{\sqrt{2}-1}\frac{\frac{1-u^2}{1+u^2}}{\frac{1-u^2}{1+u^2}+t\frac{2u}{1+u^2}}\frac{2du}{1+u^2}$$$$=2{\int }_0^{\sqrt{2}-1}\frac{1-u^2}{(1+u^2)(1-u^2+2tu)}du$$$$=2{\int }_0^{\sqrt{2}-1}\frac{u^2-1}{(u^2+1)(u^2-2tu-1)}duletdecompose$$$$F\left(u\right)=\frac{u^2-1}{(u^2+1)(u^2-2tu-1)}rootsof{u}^2-2tu-1=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=t^2+1\Rightarrow u_1=t+\sqrt{1+t^2}and{u}_2=t-\sqrt{1+t^2}\Rightarrow$$$$F\left(u\right)=\frac{a}{u-u_1}+\frac{b}{u-u_2}+\frac{cu+d}{1+u^2}....becontinued...$$

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