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Evaluate-1-0-1-dx-1-x-1-x-2-2-0-2-ln-1-2x-1-x-2-3-0-pi-x-1-sin-3-x-3picosx-4sinx-sin-2-x-4-dx-4-0-pi-x-2-cos-2-x-xsinx-cosx-1-




Question Number 54074 by rahul 19 last updated on 28/Jan/19
Evaluate :  1)      ∫_0 ^( 1) (dx/( (√(1+x))+(√(1−x))+2))     2)     ∫_0 ^( 2) ((ln(1+2x))/(1+x^2 ))  3)   ∫_0 ^( π) (x/( (√(1+sin^3 x))))((3πcosx+4sinx)sin^2 x+4)dx  4)  ∫_0 ^( π)  ((x^2 cos^2 x−xsinx−cosx−1)/((1+xsinx)^2 )) dx.
$${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right)\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}+\mathrm{2}}\: \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right) \\ $$$$\:\int_{\mathrm{0}} ^{\:\pi} \frac{{x}}{\:\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{3}} {x}}}\left(\left(\mathrm{3}\pi\mathrm{cos}{x}+\mathrm{4sin}{x}\right)\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{4}\right){dx} \\ $$$$\left.\mathrm{4}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:\frac{{x}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}−{x}\mathrm{sin}{x}−\mathrm{cos}{x}−\mathrm{1}}{\left(\mathrm{1}+{x}\mathrm{sin}{x}\right)^{\mathrm{2}} }\:{dx}. \\ $$
Commented by maxmathsup by imad last updated on 28/Jan/19
1) let I =∫_0 ^1     (dx/( (√(1+x)) +(√(1−x)) +2))  changement x=cos θ  give   I = −∫_0 ^(π/2)     ((−sinθ dθ)/( (√2)cos((θ/2))+(√2)sin((θ/2)) +2)) =(1/( (√2))) ∫_0 ^(π/2)    ((sinθ)/(cos((θ/2)) +sin((θ/2))+(√2)))dθ  =_((θ/2)=t)    (1/( (√2))) ∫_0 ^(π/4)    ((sin(2t))/(cost +sint +(√2))) (2)dt =(√2)∫_0 ^(π/4)   ((sin(2t))/(cost +sint +(√2))) dt  =2(√2)∫_0 ^(π/4)   ((sint cost)/(cost +sint +(√2))) dt =_(tan((t/2)) =x)     2(√2)∫_0 ^((√2)−1)     (((2x(1−x^2 ))/((1+x^2 )^2 ))/(((1−x^2 )/(1+x^2 )) +((2x)/(1+x^2 )) +(√2))) ((2dx)/(1+x^2 ))  =8(√2)∫_0 ^((√2)−1)       ((x(1−x^2 ))/((1+x^2 )^2 (1−x^2  +2x+(√2)(1+x^2 ))))dx  =8(√2)∫_0 ^((√2)−1)    ((x(1−x^2 ))/((1+x^2 )^2 (((√2)−1)x^2  +2x +1+(√2))))dx let decompose  F(x) =((x(1−x^2 ))/((x^2  +1)^2 { ((√2)−1)x^2  +2x +1+(√2)}))  roots of ((√2)−1)x^2  +2x +1+(√2)  Δ =4−4((√2)−1)((√2)+1) =0 ⇒one root x_0 =−(b/(2a)) =−(2/(2((√2)−1))) =((−1)/( (√2)−1)) ⇒  ((√2)−1)x^2  +2x +1+(√2)=((√2)−1)(x−x_0 )^2  =((√2)−1)(x+(1/( (√2)−1)))^2   =((√2)−1)(x +(√2)+1)^2  ⇒  F(x) =(a/(x+(√2) +1)) +(b/((x+(√2) +1)^2 )) +((cx +d)/(x^2  +1)) + ((ex +f)/((x^2  +1)^2 ))  b=lim_(x→−(1+(√2)))     (x+(√2)+1)^2 F(x)....after finding the coefficients  we get ∫ F(x)dx =aln∣x+(√2)+1∣−(b/((x+(√2) +1))) +∫  ((cx+d)/(x^2  +1))dx +∫  ((ex +f)/((x^2  +1)^2 ))dx  ...be continued...
$$\left.\mathrm{1}\right)\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}\:+\mathrm{2}}\:\:{changement}\:{x}={cos}\:\theta\:\:{give}\: \\ $$$${I}\:=\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{−{sin}\theta\:{d}\theta}{\:\sqrt{\mathrm{2}}{cos}\left(\frac{\theta}{\mathrm{2}}\right)+\sqrt{\mathrm{2}}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\:+\mathrm{2}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}\theta}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)\:+{sin}\left(\frac{\theta}{\mathrm{2}}\right)+\sqrt{\mathrm{2}}}{d}\theta \\ $$$$=_{\frac{\theta}{\mathrm{2}}={t}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{{cost}\:+{sint}\:+\sqrt{\mathrm{2}}}\:\left(\mathrm{2}\right){dt}\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{{cost}\:+{sint}\:+\sqrt{\mathrm{2}}}\:{dt} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sint}\:{cost}}{{cost}\:+{sint}\:+\sqrt{\mathrm{2}}}\:{dt}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={x}} \:\:\:\:\mathrm{2}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\frac{\mathrm{2}{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:+\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\sqrt{\mathrm{2}}}\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\mathrm{8}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\:\:\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\sqrt{\mathrm{2}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)}{dx} \\ $$$$=\mathrm{8}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}+\sqrt{\mathrm{2}}\right)}{dx}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)\:=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \left\{\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}+\sqrt{\mathrm{2}}\right\}} \\ $$$${roots}\:{of}\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\Delta\:=\mathrm{4}−\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\:=\mathrm{0}\:\Rightarrow{one}\:{root}\:{x}_{\mathrm{0}} =−\frac{{b}}{\mathrm{2}{a}}\:=−\frac{\mathrm{2}}{\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}\:=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\:\Rightarrow \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}+\sqrt{\mathrm{2}}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left({x}−{x}_{\mathrm{0}} \right)^{\mathrm{2}} \:=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left({x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left({x}\:+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}+\sqrt{\mathrm{2}}\:+\mathrm{1}}\:+\frac{{b}}{\left({x}+\sqrt{\mathrm{2}}\:+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{cx}\:+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\:\frac{{ex}\:+{f}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:\:\left({x}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)….{after}\:{finding}\:{the}\:{coefficients} \\ $$$${we}\:{get}\:\int\:{F}\left({x}\right){dx}\:={aln}\mid{x}+\sqrt{\mathrm{2}}+\mathrm{1}\mid−\frac{{b}}{\left({x}+\sqrt{\mathrm{2}}\:+\mathrm{1}\right)}\:+\int\:\:\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:+\int\:\:\frac{{ex}\:+{f}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$…{be}\:{continued}… \\ $$
Commented by maxmathsup by imad last updated on 28/Jan/19
2) let f(t) =∫_0 ^2  ((ln(1+tx))/(1+x^2 )) dx  with t>0 ⇒  f^′ (t) =∫_0 ^2   (x/((1+tx)(1+x^2 )))dx =_(tx =u)   ∫_0 ^(2t)    (u/(t(1+u)(1+(u^2 /t^2 )))) (du/t)  =∫_0 ^(2t)    ((udu)/((u+1)(u^2  +t^2 )))  let decompose F(u) = (u/((u+1)(u^2  +t^2 ))) ⇒  F(u) =(a/(u+1)) +((bu +c)/(u^2  +t^2 ))  a =lim_(u→−1) (u+1)F(u) =−(1/(t^2  +1))  lim_(u→+∞) u F(u) =0 =a+b ⇒b =(1/(t^2  +1))  F(0) =0 =a +(c/t^2 ) ⇒c =−at^2  =(t^2 /(t^(2 )  +1)) ⇒  F(u) =−(1/((t^2  +1)(u+1))) +(((1/(t^2  +1 ))u +(t^2 /(t^2  +1)))/(u^2  +t^2 ))  =(1/(t^2  +1)){ −(1/(u+1)) +((u +t^2 )/(u^2  +t^2 ))} ⇒∫_0 ^(2t)   F(u)du  =(1/(t^2  +1)) {−∫_0 ^(2t)  (du/(u+1)) +∫_0 ^(2t)   ((u+t^2 )/(u^2  +t^2 )) du}  =(1/(t^2  +1)){−[ln∣u+1∣]_0 ^(2t)  + (1/2)[ln(u^2  +t^2 )]_0 ^(2t)   +2t^2  ∫_0 ^(2t)    (du/(u^2  +t^2 ))}  =(1/(t^2  +1)){ −ln(1+2t) +(1/2)( ln(5t^2 )−ln(t^2 )) +2t^2  ∫_0 ^(2t)   (du/(u^2  +t^2 ))}  =−((ln(1+2t))/(t^2  +1)) +((ln(5))/(2(t^2  +1))) +((2t^2 )/(t^2  +1)) ∫_0 ^(2t)   (du/(u^2  +t^2 )) but  ∫_0 ^(2t)    (du/(u^2  +t^2 )) =_(u =tα)    ∫_0 ^2    ((tdα)/(t^2 (1+α^2 ))) =(1/t) ∫_0 ^2   (dα/(1+α^2 )) =((arctan(2))/t) ⇒  f^′ (t) =−((ln(1+2t))/(t^2  +1)) +((ln(5))/(2(t^2  +1))) +((2arctan(2)t)/(t^2  +1)) ⇒  f(t) =−∫_0 ^t   ((ln(1+2x))/(1+x^2 ))dx +((ln(5))/2) ∫_0 ^t  (dx/(1+x^2 )) + arctan(2) ∫_0 ^t   ((2x)/(1+x^2 ))dx +c   f(t) =−∫_0 ^t   ((ln(1+2x))/(1+x^2 ))dx +((ln(5))/2) arctan(t) +arctan(2)ln(1+t^2 ) +c  f(0)=0 ⇒c =0 ⇒  f(t) =−∫_0 ^t   ((ln(1+2x))/(1+x^2 ))dx +((ln(5))/2) arctan(t)  +arctan(2)ln(1+t^2 ) =∫_0 ^2  ((ln(1+tx))/(1+x^2 ))dx  we have f(2) =−∫_0 ^2   ((ln(1+2x))/(1+x^2 )) dx +((ln(5))/2) arctan(2) +ln(5)arctan(2)  =∫_0 ^2  ((ln(1+2x))/(1+x^2 ))dx ⇒ 2 ∫_0 ^2   ((ln(1+2x))/(1+x^2 ))dx =(3/2)ln(5) arctan(2)  ⇒  ∫_0 ^2    ((ln(1+2x))/(1+x^2 ))dx =(3/4)ln(5)arctan(2) .i
$$\left.\mathrm{2}\right)\:{let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{ln}\left(\mathrm{1}+{tx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\:{with}\:{t}>\mathrm{0}\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{{x}}{\left(\mathrm{1}+{tx}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:=_{{tx}\:={u}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:\:\frac{{u}}{{t}\left(\mathrm{1}+{u}\right)\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{t}^{\mathrm{2}} }\right)}\:\frac{{du}}{{t}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:\:\frac{{udu}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)}\:\:{let}\:{decompose}\:{F}\left({u}\right)\:=\:\frac{{u}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{a}}{{u}+\mathrm{1}}\:+\frac{{bu}\:+{c}}{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{u}\rightarrow−\mathrm{1}} \left({u}+\mathrm{1}\right){F}\left({u}\right)\:=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${lim}_{{u}\rightarrow+\infty} {u}\:{F}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}\:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:={a}\:+\frac{{c}}{{t}^{\mathrm{2}} }\:\Rightarrow{c}\:=−{at}^{\mathrm{2}} \:=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}\:} \:+\mathrm{1}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=−\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}+\mathrm{1}\right)}\:+\frac{\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}\:}{u}\:+\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}}{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:−\frac{\mathrm{1}}{{u}+\mathrm{1}}\:+\frac{{u}\:+{t}^{\mathrm{2}} }{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }\right\}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:{F}\left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\left\{−\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\frac{{du}}{{u}+\mathrm{1}}\:+\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:\frac{{u}+{t}^{\mathrm{2}} }{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }\:{du}\right\} \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\left\{−\left[{ln}\mid{u}+\mathrm{1}\mid\right]_{\mathrm{0}} ^{\mathrm{2}{t}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:+\mathrm{2}{t}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:−{ln}\left(\mathrm{1}+\mathrm{2}{t}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\:{ln}\left(\mathrm{5}{t}^{\mathrm{2}} \right)−{ln}\left({t}^{\mathrm{2}} \right)\right)\:+\mathrm{2}{t}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }\right\} \\ $$$$=−\frac{{ln}\left(\mathrm{1}+\mathrm{2}{t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\frac{\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}{t}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }\:=_{{u}\:={t}\alpha} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\frac{{td}\alpha}{{t}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{t}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\frac{{arctan}\left(\mathrm{2}\right)}{{t}}\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=−\frac{{ln}\left(\mathrm{1}+\mathrm{2}{t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\frac{\mathrm{2}{arctan}\left(\mathrm{2}\right){t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=−\int_{\mathrm{0}} ^{{t}} \:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{t}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\:{arctan}\left(\mathrm{2}\right)\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+{c}\: \\ $$$${f}\left({t}\right)\:=−\int_{\mathrm{0}} ^{{t}} \:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}}\:{arctan}\left({t}\right)\:+{arctan}\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+{c} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{c}\:=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=−\int_{\mathrm{0}} ^{{t}} \:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}}\:{arctan}\left({t}\right)\:\:+{arctan}\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{ln}\left(\mathrm{1}+{tx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${we}\:{have}\:{f}\left(\mathrm{2}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}}\:{arctan}\left(\mathrm{2}\right)\:+{ln}\left(\mathrm{5}\right){arctan}\left(\mathrm{2}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\Rightarrow\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{5}\right)\:{arctan}\left(\mathrm{2}\right)\:\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\mathrm{5}\right){arctan}\left(\mathrm{2}\right)\:.{i} \\ $$
Commented by rahul 19 last updated on 29/Jan/19
thank you prof Abdo.
$${thank}\:{you}\:{prof}\:{Abdo}. \\ $$
Commented by rahul 19 last updated on 29/Jan/19
Everyone pls try Q.3,4 also.
$${Everyone}\:{pls}\:{try}\:{Q}.\mathrm{3},\mathrm{4}\:{also}. \\ $$
Commented by maxmathsup by imad last updated on 29/Jan/19
you are welcome sir.
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19
4)∫_0 ^π (((1+xsinx)(du/dx)−u(d/dx)(1+xsinx))/((1+xsinx)^2 ))dx  ∫_0 ^π (d/dx)((u/(1+xsinx)))dx  ∫_0 ^π d((u/(1+xsinx)))  ∣(u/(1+xsinx))∣_0 ^π   u=g(x)say  =((g(π))/(1+πsinπ))−((g(0))/(1+0×sin0))  =g(π)−g(0)  wait i have to find g(x)=u...
$$\left.\mathrm{4}\right)\int_{\mathrm{0}} ^{\pi} \frac{\left(\mathrm{1}+{xsinx}\right)\frac{{du}}{{dx}}−{u}\frac{{d}}{{dx}}\left(\mathrm{1}+{xsinx}\right)}{\left(\mathrm{1}+{xsinx}\right)^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{d}}{{dx}}\left(\frac{{u}}{\mathrm{1}+{xsinx}}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} {d}\left(\frac{{u}}{\mathrm{1}+{xsinx}}\right) \\ $$$$\mid\frac{{u}}{\mathrm{1}+{xsinx}}\mid_{\mathrm{0}} ^{\pi} \:\:{u}={g}\left({x}\right){say} \\ $$$$=\frac{{g}\left(\pi\right)}{\mathrm{1}+\pi{sin}\pi}−\frac{{g}\left(\mathrm{0}\right)}{\mathrm{1}+\mathrm{0}×{sin}\mathrm{0}} \\ $$$$={g}\left(\pi\right)−{g}\left(\mathrm{0}\right) \\ $$$${wait}\:{i}\:{have}\:{to}\:{find}\:{g}\left({x}\right)={u}… \\ $$

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