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Question Number 32708 by abdo imad last updated on 31/Mar/18

$$letgivef\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{ln(1+xtant)}{tant}dt$$$$findasimpleformoff\left(x\right)$$$$2)calculate{\int }_0^{\frac{\pi }{2}}\frac{ln(1+2tant)}{tant}dt.$$

Commented by abdo imad last updated on 03/Apr/18

$$wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{\partial }{\partial x}\left(\frac{ln(1+xtant)}{tant}\right)dt$$$$={\int }_0^{\frac{\pi }{2}}\frac{tant}{(1+xtant)tant}dt={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+xtant}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x\frac{sint}{cost}}={\int }_0^{\frac{\pi }{2}}\frac{cost}{cost+xsint}dt.ch.tan\left(\frac{t}{2}\right)=u$$$$give{f}^{,}\left(x\right)={\int }_0^1\frac{\frac{1-u^2}{1+u^2}}{\frac{1-u^2}{1+u^2}+x\frac{\left(2u\right)}{1+u^2}}\frac{2du}{1+u^2}$$$$f^{{}^{\prime }}\left(x\right)={\int }_0^1\frac{2(1-u^2)}{(1+u^2)(1-u^2+2xu)}du$$$$={\int }_0^1\frac{2(u^2-1)}{(1+u^2)(u^2-2xu-1)}duletdecompose$$$$F\left(u\right)=\frac{2(u^2-1)}{(1+u^2)(u^2-2xu-1)}$$$$u^2-2xu-1\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}=x^2+1\Rightarrow u_1=x+\sqrt{1+x^2}$$$$u_2=x-\sqrt{1+x^2}$$$$F\left(u\right)=\frac{a}{u-u_1}+\frac{b}{u-u_2}+\frac{cu+d}{u^2+1}=\frac{2(u^2-1)}{(u-u_1)(u-u_2)(u^2+1)}$$$$a=\frac{2(u_1^2-1)}{(u_1-u_2)(u_1^2+1)}=\frac{2({(x+\sqrt{1+x^2})}^2-1)}{2\sqrt{1+x^2({(x+\sqrt{1+x^2})}^2+1)}}$$$$b=\frac{2(u_2^2-1)}{(u_2-u_1)(u_2^2+1)}=\frac{2({(x-\sqrt{1+x^2})}^2-1)}{-2\sqrt{2+x^2((x-\sqrt{{1+x^2)}^2}+1)}}$$$$becontinued....$$

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