Question Number 175987 by stelor last updated on 10/Sep/22
$${please}\:{calculate} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\left({tanx}\right)^{\sqrt{\mathrm{2}}} } \\ $$$${J}=\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}} \\ $$$${K}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{3}{tanx}+\mathrm{2}} \\ $$
Answered by Ar Brandon last updated on 10/Sep/22
![J=∫_0 ^π (dx/(a^2 cos^2 x+sin^2 x))=∫_0 ^π ((sec^2 x)/(a^2 +tan^2 x))dx =2∫_0 ^(π/2) ((d(tanx))/(a^2 +tan^2 x))=(2/a)[arctan(((tanx)/a))]_0 ^(π/2) =(π/a)](Q175988.png)
$${J}=\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} {x}}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{tan}{x}\right)}{{a}^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} {x}}=\frac{\mathrm{2}}{{a}}\left[\mathrm{arctan}\left(\frac{\mathrm{tan}{x}}{{a}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{{a}} \\ $$
Commented by stelor last updated on 10/Sep/22
$${Thanks} \\ $$
Answered by Ar Brandon last updated on 10/Sep/22
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\left(\mathrm{tan}{x}\right)^{\sqrt{\mathrm{2}}} }=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\mathrm{cos}{x}\right)^{\sqrt{\mathrm{2}}} }{\left(\mathrm{sin}{x}\right)^{\sqrt{\mathrm{2}}} +\left(\mathrm{cos}{x}\right)^{\sqrt{\mathrm{2}}} }{dx}\:\:...\left({i}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\mathrm{sin}{x}\right)^{\sqrt{\mathrm{2}}} }{\left(\mathrm{sin}{x}\right)^{\sqrt{\mathrm{2}}} +\left(\mathrm{cos}{x}\right)^{\sqrt{\mathrm{2}}} }{dx}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)\:+\:\left({ii}\right)\:\Rightarrow \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\mathrm{sin}{x}\right)^{\sqrt{\mathrm{2}}} +\left(\mathrm{cos}{x}\right)^{\sqrt{\mathrm{2}}} }{\left(\mathrm{sin}{x}\right)^{\sqrt{\mathrm{2}}} +\left(\mathrm{cos}{x}\right)^{\sqrt{\mathrm{2}}} }{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{I}=\frac{\pi}{\mathrm{4}} \\ $$
Answered by Ar Brandon last updated on 10/Sep/22
Commented by Ar Brandon last updated on 10/Sep/22
$${K}\approx\mathrm{0},\mathrm{335236} \\ $$
Commented by Ar Brandon last updated on 11/Sep/22
![K=∫_0 ^(π/2) (dx/(3tanx+2))=∫_0 ^1 (2/(3(((2t)/(1−t^2 )))+2))∙(dt/(1+t^2 ))=∫_0 ^1 ((t^2 −1)/(2t^2 −6t−2))∙((2dt)/(t^2 +1)) =∫_0 ^1 ((t^2 −1)/((t^2 −3t−1)(t^2 +1)))dt=(1/(13))∫_0 ^1 (((6t−9)/(t^2 −3t−1))−((6t−4)/(t^2 +1)))dt −−−−−−−−−−−−−−−−−−−−−−−−−−− ((at+b)/(t^2 −3t−1))+((ct+d)/(t^2 +1))=(((at+b)(t^2 +1)+(ct+d)(t^2 −3t−1))/((t^2 −3t−1)(t^2 +1))) a+c=^t^3 0, b−3c+d=1, a−c−3d=0, b−d=−1 lim_(t→i) ⇒(ci+d)(−2−3i)=(−2d+3c)−i(2c+3d)=−2 ⇒c=−(6/(13)), d=(4/(13)), ⇒b=−(9/(13)), a=(6/(13)) −−−−−−−−−−−−−−−−−−−−−−−−−−− =(1/(13))∫_0 ^1 ((3(2t−3))/(t^2 −3t−1))dt−(3/(13))∫_0 ^1 ((2t)/(t^2 +1))dt+(4/(13))∫_0 ^1 (1/(t^2 +1))dt =(3/(13))[ln∣t^2 −3t−1∣]_0 ^1 −(3/(13))[ln(t^2 +1)]_0 ^1 +(4/(13))[arctan(t)]_0 ^1 =(3/(13))[ln∣((t^2 −3t−1)/(t^2 +1))∣]_0 ^1 +(4/(13))((π/4))=(1/(13))ln((3/2))^3 +(π/(13)) =(1/(13))ln(((27)/8))+(π/(13))=(1/(13))(π+ln(((27)/8)))≈0,33523](Q176010.png)
$${K}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{3tan}{x}+\mathrm{2}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}\left(\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\right)+\mathrm{2}}\centerdot\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{2}}\centerdot\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} −\mathrm{1}}{\left({t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}=\frac{\mathrm{1}}{\mathrm{13}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{6}{t}−\mathrm{9}}{{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}}−\frac{\mathrm{6}{t}−\mathrm{4}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\frac{{at}+{b}}{{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}}+\frac{{ct}+{d}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\left({at}+{b}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)+\left({ct}+{d}\right)\left({t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}\right)}{\left({t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${a}+{c}\overset{{t}^{\mathrm{3}} } {=}\mathrm{0},\:{b}−\mathrm{3}{c}+{d}=\mathrm{1},\:{a}−{c}−\mathrm{3}{d}=\mathrm{0},\:{b}−{d}=−\mathrm{1} \\ $$$$\underset{{t}\rightarrow{i}} {\mathrm{lim}}\:\Rightarrow\left({ci}+{d}\right)\left(−\mathrm{2}−\mathrm{3}{i}\right)=\left(−\mathrm{2}{d}+\mathrm{3}{c}\right)−{i}\left(\mathrm{2}{c}+\mathrm{3}{d}\right)=−\mathrm{2} \\ $$$$\Rightarrow{c}=−\frac{\mathrm{6}}{\mathrm{13}},\:{d}=\frac{\mathrm{4}}{\mathrm{13}},\:\Rightarrow{b}=−\frac{\mathrm{9}}{\mathrm{13}},\:{a}=\frac{\mathrm{6}}{\mathrm{13}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{13}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{3}\left(\mathrm{2}{t}−\mathrm{3}\right)}{{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}}{dt}−\frac{\mathrm{3}}{\mathrm{13}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\frac{\mathrm{4}}{\mathrm{13}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\:\:\:\:=\frac{\mathrm{3}}{\mathrm{13}}\left[\mathrm{ln}\mid{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{13}}\left[\mathrm{ln}\left({t}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{4}}{\mathrm{13}}\left[\mathrm{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:=\frac{\mathrm{3}}{\mathrm{13}}\left[\mathrm{ln}\mid\frac{{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{4}}{\mathrm{13}}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{13}}\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} +\frac{\pi}{\mathrm{13}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{13}}\mathrm{ln}\left(\frac{\mathrm{27}}{\mathrm{8}}\right)+\frac{\pi}{\mathrm{13}}=\frac{\mathrm{1}}{\mathrm{13}}\left(\pi+\mathrm{ln}\left(\frac{\mathrm{27}}{\mathrm{8}}\right)\right)\approx\mathrm{0},\mathrm{33523} \\ $$
Answered by a.lgnaoui last updated on 10/Sep/22
![k=∫_0 ^(π/2) (dx/(3tanx+2)) posons t=tanx x=arctant dt=(1+t^2 )dx dx=(dt/(1+t^2 )) k=∫_0 ^∞ (dt/((3t+2)(1+t^2 )))=∫_0 ^∞ (((−3t+2)/(13(1+t^2 )))+(9/(13(3t+2))))dt =−(3/(26))∫((2t−(2/3))/(1+t^2 ))dt+(9/(13))∫(dt/(3t+2)) =−(3/(26))[log(1+t^2 )]_0 ^∞ +(1/(13))∫_0 ^∞ (dt/(1+t^2 ))+(9/(13))[log(3t+2)]_0 ^∞ =((−3)/(26))[log(1+t^2 )]_0 ^∞ +(1/(13))[arctan(t)]_0 ^∞ +(9/(13))[log(3t+2)]_0 ^∞ ((−1)/(13))[log(1+t^2 )]_0 ^∞ +(9/(13))[log(3t+2)]_0 ^∞ +(π/(26)) =log[((((3t+2)^(9/(13)) )/((1+t^2 )^(1/(13)) )))]_0 ^∞ +(π/(26)) t=0 (((3t+2)^(9/(13)) )/((1+t)^(1/(13)) )) =2^(9/(13)) t→∞ ((3t^(9/(13)) (1+(2/t))^(9/(13)) )/(t^(1/(13)) (1+(1/t))^(1/(13)) ))=3^(9/(13)) t^(8/(13)) k=(9/(13))log((3/2))+(π/(26))+(8/(13)) lim_(t→∞) (logt) k=+∞](Q176006.png)
$$ \\ $$$${k}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{3}{tanx}+\mathrm{2}}\:\:\:\:\:\:{posons}\:{t}={tanx}\:\:\:{x}={arctant} \\ $$$${dt}=\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dx}\:\:\:\:\:\:\:{dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${k}=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left(\mathrm{3}{t}+\mathrm{2}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\int_{\mathrm{0}} ^{\infty} \left(\frac{−\mathrm{3}{t}+\mathrm{2}}{\mathrm{13}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}+\frac{\mathrm{9}}{\mathrm{13}\left(\mathrm{3}{t}+\mathrm{2}\right)}\right){dt} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{26}}\int\frac{\mathrm{2}{t}−\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}+\frac{\mathrm{9}}{\mathrm{13}}\int\frac{{dt}}{\mathrm{3}{t}+\mathrm{2}} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{26}}\left[\mathrm{log}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{13}}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{13}}\left[\mathrm{log}\left(\mathrm{3t}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{−\mathrm{3}}{\mathrm{26}}\left[\mathrm{log}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{13}}\left[{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{9}}{\mathrm{13}}\left[\mathrm{log}\left(\mathrm{3t}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$ \\ $$$$\frac{−\mathrm{1}}{\mathrm{13}}\left[\mathrm{log}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{9}}{\mathrm{13}}\left[\mathrm{log}\left(\mathrm{3t}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\pi}{\mathrm{26}} \\ $$$$=\mathrm{log}\left[\left(\frac{\left(\mathrm{3t}+\mathrm{2}\right)^{\frac{\mathrm{9}}{\mathrm{13}}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{13}}} }\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\pi}{\mathrm{26}} \\ $$$$\mathrm{t}=\mathrm{0}\:\:\:\:\frac{\left(\mathrm{3t}+\mathrm{2}\right)^{\frac{\mathrm{9}}{\mathrm{13}}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{13}}} }\:=\mathrm{2}^{\frac{\mathrm{9}}{\mathrm{13}}} \:\:\:\: \\ $$$${t}\rightarrow\infty\:\:\:\:\frac{\mathrm{3}{t}^{\frac{\mathrm{9}}{\mathrm{13}}} \left(\mathrm{1}+\frac{\mathrm{2}}{{t}}\right)^{\frac{\mathrm{9}}{\mathrm{13}}} }{{t}^{\frac{\mathrm{1}}{\mathrm{13}}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)^{\frac{\mathrm{1}}{\mathrm{13}}} }=\mathrm{3}^{\frac{\mathrm{9}}{\mathrm{13}}} {t}^{\frac{\mathrm{8}}{\mathrm{13}}} \\ $$$${k}=\frac{\mathrm{9}}{\mathrm{13}}\mathrm{log}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{26}}+\frac{\mathrm{8}}{\mathrm{13}}\:\mathrm{lim}_{\mathrm{t}\rightarrow\infty} \left(\mathrm{logt}\right) \\ $$$${k}=+\infty \\ $$$$ \\ $$