Calcul ∫^( (π/2)) _( 0) t(√(tan(t))) dt


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Question Number 195895 by Erico last updated on 12/Aug/23

$Calcul {\int_{0}}^{\frac{π}{2}} t \sqrt{\tan (t)} dt$
	Answered by witcher3 last updated on 13/Aug/23
	$∫_0 ^∞ ((tan^(−1) (t).(√t))/(1+t^2 )) =∫_0 ^∞ ((√t)/(1+t^2 )).∫_0 ^1 ((tdx)/(1+(xt)^2 )).dx =∫_0 ^1 ∫_0 ^∞ (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))dtdx ∫_0 ^∞ (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))dt=A ∫_(−∞) ^0 (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))=∫_0 ^∞ ((−it^(3/2) )/((1+t^2 )(1+x^2 t^2 )))dt=A ∫_C (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))=(((i)^(3/2) )/((2i)(1−x^2 ))).+(((i)^(3/2) x^2 )/(x^(3/2) (x^2 −1).2ix)) =(i^(1/2) /(2(1−x^2 )))(1−(1/( (√x)))) =((i^(3/2) π)/( (√x)(1+x)(1+(√x))))=((−πe^((i3π)/4) )/( (√x)(1+(√x))(1+x))). ∫_0 ^1 ((π(√2))/((1+t)(1+t^2 ))). =∫_0 ^1 ((π(√2)[1+t^2 +(1−t)(1+t))/(2(1+t)(1+t^2 ))) =(π/( (√2))){∫_0 ^1 (dt/(1+t))+∫_0 ^1 (dt/(1+t^2 ))−∫_0 ^1 ((tdt)/(1+t^2 ))} =(π/( (√2))){ln(2)+(π/4)−(1/2)ln(2))} =(π/( (√2)))[((ln(2))/2)+(π/4)]=∫_0 ^(π/2) x(√(tan (x)))dx$
	$\int_{0}^{\infty} \frac{\tan^{- 1} (t) . \sqrt{t}}{1 + t^{2}}$ $= \int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} . \int_{0}^{1} \frac{tdx}{1 + {(xt)}^{2}} . dx$ $= \int_{0}^{1} \int_{0}^{\infty} \frac{t^{\frac{3}{2}}}{(1 + t^{2}) (1 + x^{2} t^{2})} dtdx$ $\int_{0}^{\infty} \frac{t^{\frac{3}{2}}}{(1 + t^{2}) (1 + x^{2} t^{2})} dt = A$ $\int_{- \infty}^{0} \frac{t^{\frac{3}{2}}}{(1 + t^{2}) (1 + x^{2} t^{2})} = \int_{0}^{\infty} \frac{- {it}^{\frac{3}{2}}}{(1 + t^{2}) (1 + x^{2} t^{2})} dt = A$ $\int_{C} \frac{t^{\frac{3}{2}}}{(1 + t^{2}) (1 + x^{2} t^{2})} = \frac{{(i)}^{\frac{3}{2}}}{(2 i) (1 - x^{2})} . + \frac{{(i)}^{\frac{3}{2}} x^{2}}{x^{\frac{3}{2}} (x^{2} - 1) . 2 ix}$ $= \frac{i^{\frac{1}{2}}}{2 (1 - x^{2})} (1 - \frac{1}{\sqrt{x}})$ $= \frac{i^{\frac{3}{2}} π}{\sqrt{x} (1 + x) (1 + \sqrt{x})} = \frac{- π e^{\frac{i 3 π}{4}}}{\sqrt{x} (1 + \sqrt{x}) (1 + x)} .$ $\int_{0}^{1} \frac{π \sqrt{2}}{(1 + t) (1 + t^{2})} .$ $= \int_{0}^{1} \frac{π \sqrt{2} [1 + t^{2} + (1 - t) (1 + t)}{2 (1 + t) (1 + t^{2})}$ $= \frac{π}{\sqrt{2}} {\int_{0}^{1} \frac{dt}{1 + t} + \int_{0}^{1} \frac{dt}{1 + t^{2}} - \int_{0}^{1} \frac{tdt}{1 + t^{2}}}$ $= \frac{π}{\sqrt{2}} {\ln (2) + \frac{π}{4} - \frac{1}{2} \ln (2))}$ $= \frac{π}{\sqrt{2}} [\frac{\ln (2)}{2} + \frac{π}{4}] = \int_{0}^{\frac{π}{2}} x \sqrt{\tan (x)} dx$
	Commented by Rodier97 last updated on 17/Aug/23

	$Hi . . . . . i^{'} m having trouble understanding how you did$ $the second line . . . i^{'} m still trying to notice that$ $\tan^{- 1} (t) = \int_{0}^{1} \frac{1}{1 + {(xt)}^{2}} dx$ $but where does^{″} {tdx}^{″} come from in your term$ $\int_{0}^{1} \frac{tdx}{1 + {(xt)}^{2}} dx ? ? ?$ $please!$
	Commented by witcher3 last updated on 20/Aug/23

	$\int_{0}^{1} \frac{tdx}{1 + {(xt)}^{2}} = a; xt = y \Rightarrow dy = tdx \Rightarrow a = \int_{0}^{t} \frac{dy}{1 + y^{2}} = \tan^{- 1} (t)$
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