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Question Number 97569 by  M±th+et+s last updated on 08/Jun/20
∫_0 ^(π/4) (√(tan(x)))(√(1−tan(x))) dx
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{{tan}\left({x}\right)}\sqrt{\mathrm{1}−{tan}\left({x}\right)}\:{dx} \\ $$
Answered by MJS last updated on 08/Jun/20
use thus: t=((√(1−tan x))/( (√(tan x)))) → dx=−2cos^2 x(√(tan^3 x))(√(1−tan x))dt ⇒ 2∫(t^2 /((t^2 +1)(t^4 +2t^2 +2)))dt now decompose...
$$\mathrm{use}\:\mathrm{thus}: \\ $$$${t}=\frac{\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}{\:\sqrt{\mathrm{tan}\:{x}}}\:\rightarrow\:{dx}=−\mathrm{2cos}^{\mathrm{2}} \:{x}\sqrt{\mathrm{tan}^{\mathrm{3}} \:{x}}\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}{dt} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}\right)}{dt} \\ $$$$\mathrm{now}\:\mathrm{decompose}… \\ $$
Commented by MJS last updated on 08/Jun/20
t^4 +2t^2 +2=(t^2 −(√(2(√2)−2))t+(√2))(t^2 +(√(2(√2)−2))t+(√2))
$${t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}=\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}{t}+\sqrt{\mathrm{2}}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}{t}+\sqrt{\mathrm{2}}\right) \\ $$
Commented by  M±th+et+s last updated on 10/Jun/20
thank you sir mjs (t^2 /((t^4 +2t^2 +2)(1+t^2 )))=((t^2 +2)/(t^4 +2t^2 +2))−(1/(1+t^2 )) simplifying ∫_0 ^(π/4) (√(tan(x)))(√(1−tan(x)))dx=2∫((t^2 +2)/(t^4 +2t^2 +2))dt−π with the simultanus t=((2)^(1/4) /u) and t=(2)^(1/4) u and averaging,the integral reduced to: ((1+(√2))/( (2)^(1/4) ))∫_0 ^∞ (((u^2 +1))/(u^4 +(2)^(1/4) u^2 +1))du v=u−(1/u) ∫_(−∞) ^∞ (dv/(v^2 +2+(√2)))dv−π=π((√((1+(√2))/2))−1)
$${thank}\:{you}\:{sir}\:{mjs} \\ $$$$\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{{t}^{\mathrm{2}} +\mathrm{2}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$ \\ $$$${simplifying} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{{tan}\left({x}\right)}\sqrt{\mathrm{1}−{tan}\left({x}\right)}{dx}=\mathrm{2}\int\frac{{t}^{\mathrm{2}} +\mathrm{2}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}{dt}−\pi \\ $$$${with}\:{the}\:{simultanus}\:{t}=\frac{\sqrt[{\mathrm{4}}]{\mathrm{2}}}{{u}}\:{and}\:{t}=\sqrt[{\mathrm{4}}]{\mathrm{2}}{u} \\ $$$${and}\:{averaging},{the}\:{integral}\:{reduced}\:{to}: \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{{u}^{\mathrm{4}} +\sqrt[{\mathrm{4}}]{\mathrm{2}}{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$${v}={u}−\frac{\mathrm{1}}{{u}} \\ $$$$\int_{−\infty} ^{\infty} \frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{2}+\sqrt{\mathrm{2}}}{dv}−\pi=\pi\left(\sqrt{\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}}−\mathrm{1}\right) \\ $$

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