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Question Number 32039 by abdo imad last updated on 18/Mar/18

$$a>-1calculate{\int }_0^{\frac{\pi }{2}}\frac{dt}{1+a{tan}^{2}t}.$$$$2)find{\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{{(1+{atan}^{2}t)}^2}dt$$$$3)findthevalueof{\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{{(1+2{tan}^{2}t)}^2}dt.$$

Commented by abdo imad last updated on 22/Mar/18

$$letputf\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+{atan}^{2}t}$$$$f\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+a\frac{{sin}^{2}t}{{cos}^{2}t}}={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{2}t}{{cos}^{2}t+{asin}^{2}t}dt$$$$f\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{\frac{1+cos\left(2t\right)}{2}}{\frac{1+cos(2t}{2}+a\frac{1-cos\left(2t\right)}{2}}dt$$$$={\int }_0^{\frac{\pi }{2}}\frac{1+cos\left(2t\right)}{1+cos\left(2t\right)+a(1-cos\left(2t\right))}dt$$$$={\int }_0^{\frac{\pi }{2}}\frac{1+cos\left(2t\right)}{1+a+(1-a)cos\left(2t\right)}thech.2t=ugive$$$$f\left(a\right)=\frac{1}{2}{\int }_0^{\pi }\frac{1+cos\left(u\right)}{1+a+(1-a)cos\left(u\right)}duch.tan\left(\frac{u}{2}\right)=xgive$$$$f\left(a\right)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{1-x^2}{1+x^2}}{1+a+(1-a)\frac{1-x^2}{1+x^2}}\frac{2dx}{1+x^2}$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{2}{(1+x^2)((1+a)(1+x^2)+(1-a)(1-x^2)}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{2dx}{(1+x^2)(1+a+(1+a)x^2+(1-a)+(a-1)x^2)}$$$$={\int }_0^{\mathrm{\infty }}\frac{2dx}{(1+x^2)(2+2{ax}^2)}={\int }_0^{\mathrm{\infty }}\frac{dx}{(1+x^2)(1+{ax}^2)}$$$$=\frac{1}{1-a}{\int }_0^{\mathrm{\infty }}(\frac{1}{1+x^2}-\frac{a}{1+{ax}^2})dx$$$$f\left(a\right)=\frac{\pi }{2(1-a)}-\frac{a}{1-a}{\int }_0^{\mathrm{\infty }}\frac{dx}{1+{ax}^2}butwehave$$$$bych.\sqrt{a}x=\alpha (wetakea>0)$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+{ax}^2}={\int }_0^{\mathrm{\infty }}\frac{1}{1+{\alpha }^2}\frac{d\alpha }{\sqrt{a}}=\frac{\pi }{2\sqrt{a}}\Rightarrow$$$$f\left(a\right)=\frac{\pi }{2(1-a)}-\frac{a}{1-a}\frac{\pi }{2\sqrt{a}}$$$$f\left(a\right)=\frac{\pi }{2(1-a)}(1-\sqrt{a})\Rightarrow f\left(a\right)=\frac{\pi }{2(1+\sqrt{a})}.$$

Commented by prof Abdo imad last updated on 22/Mar/18

$$2)wehaveprovedthat$$$${\int }_0^{\frac{\pi }{2}}\frac{dt}{1+{atan}^{2}t}=f\left(a\right)=\frac{\pi }{2(1+\sqrt{a})}letderivate$$$$f^{{}^{\prime }}\left(a\right)=-{\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{{(1+a{tan}^{2}t)}^2}dt\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{{(1+a{tan}^{2}t)}^2}dt=-f^{{}^{\prime }}\left(a\right)but$$$$f^{{}^{\prime }}\left(a\right)=-\frac{\pi }{2}\left(\frac{{1+\sqrt{a})}^{{}^{\prime }}}{{(1+\sqrt{a})}^2}\right)=-\frac{\pi }{2}\frac{1}{2\sqrt{a}{(1+\sqrt{a})}^2}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{(1+{atan}^{2}t)}dt=\frac{\pi }{4\sqrt{a}{(1+\sqrt{a})}^2}$$$$3)a=2\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{(1+2{tan}^{2}t)}dt=\frac{\pi }{{4\sqrt{2(}1+\sqrt{2})}^2}.$$

Commented by abdo imad last updated on 22/Mar/18

$$a>0.$$

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