Question Number 83441 by mathmax by abdo last updated on 02/Mar/20
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}+\mathrm{2}{tanx}}{dx} \\ $$
Answered by M±th+et£s last updated on 06/Mar/20
![∫_0 ^(π/4) (√(1+2tan(x))) dx=∫_0 ^(π/4) ((1+2tan(x))/(√(1+2tan(x))))dx =∫_0 ^(π/4) ((1+2tan(x))/(1+tan^2 (x))) ((sec^2 (x))/(√(1+2tan(x)))) dx y=(√(1+2tanx)) dy =((sec^2 (x))/(√(1+2tan(x)))) ((y^2 −1)/2)=tan(x) if x=0 y=1 if x=(π/4) y=(√3) ∫_1 ^(√3) (y^2 /(1+(((y^2 −1)/2))^2 )) dy 4∫_1 ^(√3) (y^2 /(y^4 −2y^2 +5))dy=2∫_1 ^(√3) ((y^2 −(√5) +y^2 +(√5))/(y^4 −2y^2 +5))dy =2∫_1 ^(√3) ((1−((√5)/y^2 ))/(y^2 +(5/y^2 )−2))dy+2∫_1 ^(√3) ((1+(5/y^2 ))/(y^2 +(5/y^2 )−2)) dy =2∫_1 ^(√3) ((d(y+((√5)/y)))/((y+((√5)/y))^2 −2−2(√5))) + 2∫_1 ^(√3) ((d(y−((√5)/y)))/((y−((√5)/y))^2 +2(√5)−2)) =((−2)/(√(2+2(√5))))[tanh^(−1) (((y+((√5)/y))/(√(2+2(√5)))))]_1 ^(√3) +(2/(√(2(√5)−2)))[tan^(−1) (((y−((√5)/y))/(√(2(√5)−2))))]_1 ^(√3) ((−2)/(√(2(√5)+2)))[tanh^(−1) ((((√3) +((√5)/(√3)))/(√(2+2(√5)))))−tanh^(−1) (((1+(√5))/(√(2+2(√5)))))]+(2/(√(2(√5)−2)))[tan^(−1) ((((√3)−((√5)/(√3)))/(√(2(√5)−2))))−tan^(−1) (((1−(√5))/(√(2(√5)−2))))]](Q83841.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}+\mathrm{2}{tan}\left({x}\right)}\:{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+\mathrm{2}{tan}\left({x}\right)}{\sqrt{\mathrm{1}+\mathrm{2}{tan}\left({x}\right)}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+\mathrm{2}{tan}\left({x}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}\:\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\sqrt{\mathrm{1}+\mathrm{2}{tan}\left({x}\right)}}\:{dx} \\ $$$${y}=\sqrt{\mathrm{1}+\mathrm{2}{tanx}}\:{dy}\:=\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\sqrt{\mathrm{1}+\mathrm{2}{tan}\left({x}\right)}} \\ $$$$\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}={tan}\left({x}\right) \\ $$$${if}\:{x}=\mathrm{0}\:{y}=\mathrm{1} \\ $$$${if}\:{x}=\frac{\pi}{\mathrm{4}}\:{y}=\sqrt{\mathrm{3}} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{y}^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:{dy} \\ $$$$\mathrm{4}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{y}^{\mathrm{2}} }{{y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{5}}{dy}=\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{y}^{\mathrm{2}} −\sqrt{\mathrm{5}}\:+{y}^{\mathrm{2}} +\sqrt{\mathrm{5}}}{{y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{5}}{dy} \\ $$$$ \\ $$$$=\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{{y}^{\mathrm{2}} }}{{y}^{\mathrm{2}} +\frac{\mathrm{5}}{{y}^{\mathrm{2}} }−\mathrm{2}}{dy}+\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{1}+\frac{\mathrm{5}}{{y}^{\mathrm{2}} }}{{y}^{\mathrm{2}} +\frac{\mathrm{5}}{{y}^{\mathrm{2}} }−\mathrm{2}}\:{dy} \\ $$$$=\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{d}\left({y}+\frac{\sqrt{\mathrm{5}}}{{y}}\right)}{\left({y}+\frac{\sqrt{\mathrm{5}}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}−\mathrm{2}\sqrt{\mathrm{5}}}\:+\:\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{d}\left({y}−\frac{\sqrt{\mathrm{5}}}{{y}}\right)}{\left({y}−\frac{\sqrt{\mathrm{5}}}{{y}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$$ \\ $$$$=\frac{−\mathrm{2}}{\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}}\left[{tanh}^{−\mathrm{1}} \left(\frac{{y}+\frac{\sqrt{\mathrm{5}}}{{y}}}{\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}}\right)\right]_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} +\frac{\mathrm{2}}{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}}\left[{tan}^{−\mathrm{1}} \left(\frac{{y}−\frac{\sqrt{\mathrm{5}}}{{y}}}{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}}\right)\right]_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \\ $$$$\frac{−\mathrm{2}}{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{2}}}\left[{tanh}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}\:+\frac{\sqrt{\mathrm{5}}}{\sqrt{\mathrm{3}}}}{\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}}\right)−{tanh}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}}\right)\right]+\frac{\mathrm{2}}{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}}\left[{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}−\frac{\sqrt{\mathrm{5}}}{\sqrt{\mathrm{3}}}}{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}}\right)−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}}\right)\right] \\ $$$$ \\ $$