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Question Number 175987 by stelor last updated on 10/Sep/22

$$pleasecalculate$$$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\left(tanx\right)}^{\sqrt{2}}}$$$$J={\int }_0^{\pi }\frac{dx}{a^2{cos}^{2}x+{sin}^{2}x}$$$$K={\int }_0^{\frac{\pi }{2}}\frac{dx}{3tanx+2}$$

Answered by Ar Brandon last updated on 10/Sep/22

$$J={\int }_0^{\pi }\frac{dx}{a^2{\cos }^{2}x+{\sin }^{2}x}={\int }_0^{\pi }\frac{{\sec }^{2}x}{a^2+{\tan }^{2}x}dx$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{d\left(\tan x\right)}{a^2+{\tan }^{2}x}=\frac{2}{a}{\left[\arctan \left(\frac{\tan x}{a}\right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{a}$$

Commented by stelor last updated on 10/Sep/22

$$Thanks$$

Answered by Ar Brandon last updated on 10/Sep/22

$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\left(\tan x\right)}^{\sqrt{2}}}={\int }_0^{\frac{\pi }{2}}\frac{{\left(\cos x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}dx...\left(i\right)$$$$I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\sin x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}dx...\left(ii\right)$$$$\left(i\right)+\left(ii\right)\Rightarrow$$$$2I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}dx=\frac{\pi }{2}$$$$\Rightarrow I=\frac{\pi }{4}$$

Answered by Ar Brandon last updated on 10/Sep/22

Commented by Ar Brandon last updated on 10/Sep/22

$$K\approx 0,335236$$

Commented by Ar Brandon last updated on 11/Sep/22

$$K={\int }_0^{\frac{\pi }{2}}\frac{dx}{3\tan x+2}={\int }_0^1\frac{2}{3\left(\frac{2t}{1-t^2}\right)+2}\cdot \frac{dt}{1+t^2}={\int }_0^1\frac{t^2-1}{2t^2-6t-2}\cdot \frac{2dt}{t^2+1}$$$$={\int }_0^1\frac{t^2-1}{(t^2-3t-1)(t^2+1)}dt=\frac{1}{13}{\int }_0^1(\frac{6t-9}{t^2-3t-1}-\frac{6t-4}{t^2+1})dt$$$$---------------------------$$$$\frac{at+b}{t^2-3t-1}+\frac{ct+d}{t^2+1}=\frac{(at+b)(t^2+1)+(ct+d)(t^2-3t-1)}{(t^2-3t-1)(t^2+1)}$$$$a+c\stackrel{t^3}{=}0,b-3c+d=1,a-c-3d=0,b-d=-1$$$$\underset{t\to i}{\lim }\Rightarrow (ci+d)(-2-3i)=(-2d+3c)-i(2c+3d)=-2$$$$\Rightarrow c=-\frac{6}{13},d=\frac{4}{13},\Rightarrow b=-\frac{9}{13},a=\frac{6}{13}$$$$---------------------------$$$$=\frac{1}{13}{\int }_0^1\frac{3(2t-3)}{t^2-3t-1}dt-\frac{3}{13}{\int }_0^1\frac{2t}{t^2+1}dt+\frac{4}{13}{\int }_0^1\frac{1}{t^2+1}dt$$$$=\frac{3}{13}{[\ln \mid t^2-3t-1\mid ]}_0^1-\frac{3}{13}{\left[\ln (t^2+1)\right]}_0^1+\frac{4}{13}{\left[\arctan \left(t\right)\right]}_0^1$$$$=\frac{3}{13}{[\ln \mid \frac{t^2-3t-1}{t^2+1}\mid ]}_0^1+\frac{4}{13}\left(\frac{\pi }{4}\right)=\frac{1}{13}\ln {\left(\frac{3}{2}\right)}^3+\frac{\pi }{13}$$$$=\frac{1}{13}\ln \left(\frac{27}{8}\right)+\frac{\pi }{13}=\frac{1}{13}(\pi +\ln \left(\frac{27}{8}\right))\approx 0,33523$$

Answered by a.lgnaoui last updated on 10/Sep/22

$$$$$$k={\int }_0^{\frac{\pi }{2}}\frac{dx}{3tanx+2}posonst=tanxx=arctant$$$$dt=(1+t^2)dxdx=\frac{dt}{1+t^2}$$$$k={\int }_0^{\mathrm{\infty }}\frac{dt}{(3t+2)(1+t^2)}={\int }_0^{\mathrm{\infty }}(\frac{-3t+2}{13(1+t^2)}+\frac{9}{13(3t+2)})dt$$$$=-\frac{3}{26}\int \frac{2t-\frac{2}{3}}{1+t^2}dt+\frac{9}{13}\int \frac{dt}{3t+2}$$$$=-\frac{3}{26}{\left[\log (1+{\mathrm{t}}^2)\right]}_0^{\mathrm{\infty }}+\frac{1}{13}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}+\frac{9}{13}{\left[\log (3\mathrm{t}+2)\right]}_0^{\mathrm{\infty }}$$$$=\frac{-3}{26}{\left[\log (1+{\mathrm{t}}^2)\right]}_0^{\mathrm{\infty }}+\frac{1}{13}{\left[arctan\left(t\right)\right]}_0^{\mathrm{\infty }}+\frac{9}{13}{\left[\log (3\mathrm{t}+2)\right]}_0^{\mathrm{\infty }}$$$$$$$$\frac{-1}{13}{\left[\log (1+{\mathrm{t}}^2)\right]}_0^{\mathrm{\infty }}+\frac{9}{13}{\left[\log (3\mathrm{t}+2)\right]}_0^{\mathrm{\infty }}+\frac{\pi }{26}$$$$=\log {\left[\left(\frac{{(3\mathrm{t}+2)}^{\frac{9}{13}}}{{(1+{\mathrm{t}}^2)}^{\frac{1}{13}}}\right)\right]}_0^{\mathrm{\infty }}+\frac{\pi }{26}$$$$\mathrm{t}=0\frac{{(3\mathrm{t}+2)}^{\frac{9}{13}}}{{(1+\mathrm{t})}^{\frac{1}{13}}}=2^{\frac{9}{13}}$$$$t\to \mathrm{\infty }\frac{3t^{\frac{9}{13}}{(1+\frac{2}{t})}^{\frac{9}{13}}}{t^{\frac{1}{13}}{(1+\frac{1}{t})}^{\frac{1}{13}}}=3^{\frac{9}{13}}{t}^{\frac{8}{13}}$$$$k=\frac{9}{13}\log \left(\frac{3}{2}\right)+\frac{\pi }{26}+\frac{8}{13}{\lim }_{\mathrm{t}\to \mathrm{\infty }}\left(\mathrm{logt}\right)$$$$k=+\mathrm{\infty }$$$$$$

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