Question-109989

Question Number 109989 by 150505R last updated on 26/Aug/20

Commented by 150505R last updated on 26/Aug/20

$${please}\:{solve} \\ $$

Answered by maths mind last updated on 30/Aug/20

∫_b ^a f(x)dx=∫_a ^b f(a+b−x)dx ⇒∫_0 ^1 ((tg^− ((x/(x+1))))/(tg^− (((1−2x^2 +2x)/2))))dx=∫_0 ^1 ((tg^− (((1−x)/(2−x))))/(tg^− (((1−2(1−x)^2 +2(1−x))/2))))dx =∫_0 ^1 ((tg^− (((1−x)/(2−x)))dx)/(tg^− (((1−2x^2 +2x)/2))))dx⇒ 2∫_0 ^1 ((tg^− ((x/(x+1)))dx)/(tg^− (((1−2x^2 +2x)/2))))=2A=∫_0 ^1 ((tg^− ((x/(x+1)))+tg^− (((1−x)/(2−x))))/(tg^− (((1−2x^2 +2x)/2))))dx tg^− (a)+tg^− (b)=tg^− (((a+b)/(1−ab))) ⇒tg^− ((x/(x+1)))+tg^− (((1−x)/(2−x)))=tg^− ((((x/(x+1))+((1−x)/(2−x)))/(1−(x/(x+1)).((1−x)/(2−x))))) =((tg^− (((−2x^2 +2x+1)/((x+1)(2−x))).(((x+1)(2−x))/2)))/(tg^− (((1−2x^2 +2x)/2)))) ⇔2A=∫_0 ^1 ((tg^− (((1−2x^2 +2x)/2)))/(tg^− (((1−2x^2 +2x)/2))))dx=∫_0 ^1 dx=1 ⇒A=(1/2)

$$\int_{{b}} ^{{a}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tg}^{−} \left(\frac{{x}}{{x}+\mathrm{1}}\right)}{{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tg}^{−} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right)}{{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tg}^{−} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right){dx}}{{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}}\right)}{dx}\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tg}^{−} \left(\frac{{x}}{{x}+\mathrm{1}}\right)\mathrm{dx}}{\mathrm{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}}{\mathrm{2}}\right)}=\mathrm{2}{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tg}^{−} \left(\frac{{x}}{{x}+\mathrm{1}}\right)+{tg}^{−} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right)}{{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}}\right)}{dx} \\ $$$${tg}^{−} \left({a}\right)+{tg}^{−} \left({b}\right)={tg}^{−} \left(\frac{{a}+{b}}{\mathrm{1}−{ab}}\right) \\ $$$$\Rightarrow{tg}^{−} \left(\frac{{x}}{{x}+\mathrm{1}}\right)+{tg}^{−} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right)={tg}^{−} \left(\frac{\frac{{x}}{{x}+\mathrm{1}}+\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}}{\mathrm{1}−\frac{{x}}{{x}+\mathrm{1}}.\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}}\right) \\ $$$$=\frac{{tg}^{−} \left(\frac{−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\left({x}+\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}.\frac{\left({x}+\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}{\mathrm{2}}\right)}{{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}}\right)} \\ $$$$\Leftrightarrow\mathrm{2}{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}}\right)}{{tg}^{−} \left(\frac{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {dx}=\mathrm{1} \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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