∫_0 ^1 t(√((1−t)/(1+t)))dt=?


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Question Number 182726 by SANOGO last updated on 13/Dec/22

$\int_{0}^{1} t \sqrt{\frac{1 - t}{1 + t}} d t = ?$
Commented by CElcedricjunior last updated on 13/Dec/22
$∫_0 ^1 t(√((1−t)/(1+t)))dt=k posons (√((1−t)/(1+t)))=x ((1−t)/(1+t))=x^2 =>1−t=tx^2 +x^2 =>t=((1−x^2 )/(1+x^2 )) dt=((−4x)/((1+x^2 )^2 ))dx=>qd: { ((t=0)),((t=1)) :}=> { ((x=1)),((x=0)) :} k=∫_0 ^1 ((4x(1−x^2 ))/((1+x^2 )^3 ))dx=∫((−4x(x^2 +1)+8x)/((1+x^2 )^3 ))dx =∫_0 ^1 ((−4x)/((1+x^2 )^2 _ ))dx+∫_0 ^1 ((8x)/((1+x^2 )^3 ))dx =[(2/(1+x^2 ))]_0 ^1 −[(2/((1+x^2 )^2 ))]_0 ^1 =1−2−(1/2)+2 k=(1/2) ======================= ..........le celebre cedric junior........673804515 ====================================$
$\int_{0}^{1} t \sqrt{\frac{1 - t}{1 + t}} d t = k$ $p o s o n s \sqrt{\frac{1 - t}{1 + t}} = x$ $\frac{1 - t}{1 + t} = x^{2} => 1 - t = {t x}^{2} + x^{2} => t = \frac{1 - x^{2}}{1 + x^{2}}$ $d t = \frac{- 4 x}{{(1 + x^{2})}^{2}} d x => q d : {\begin{cases} t = 0 \\ t = 1 \end{cases} => {\begin{cases} x = 1 \\ x = 0 \end{cases}$ $k = \int_{0}^{1} \frac{4 x (1 - x^{2})}{{(1 + x^{2})}^{3}} d x = \int \frac{- 4 x (x^{2} + 1) + 8 x}{{(1 + x^{2})}^{3}} d x$ $Missing \left or extra \right$ $= {[\frac{2}{1 + x^{2}}]}_{0}^{1} - {[\frac{2}{{(1 + x^{2})}^{2}}]}_{0}^{1}$ $= 1 - 2 - \frac{1}{2} + 2$ $k = \frac{1}{2}$ $=======================$ $. . . . . . . . . . l e c e l e b r e c e d r i c j u n i o r . . . . . . . . 673804515$ $====================================$
Commented by ARUNG_Brandon_MBU last updated on 13/Dec/22

$k = \int_{0}^{1} \frac{- 4 x^{2} (1 - x^{2})}{{(1 + x^{2})}^{3}} d x$
Commented by ARUNG_Brandon_MBU last updated on 13/Dec/22

Commented by SANOGO last updated on 13/Dec/22

$m e r c i$
	Answered by SEKRET last updated on 13/Dec/22

	$\int_{0}^{1} t \cdot \frac{\sqrt{1 - t^{2}}}{1 + t} dt \| \begin{matrix} \sin (t) = x \\ dx = \cos (t) dt \end{matrix} \|$ $\int_{0}^{\frac{π}{2}} \sin (t) \cdot \frac{\cos (t)}{1 + \sin (t)} \cdot \cos (t) dt =$ $\int_{0}^{\frac{π}{2}} \frac{\sin (t) \cdot (1 - sint) (1 + sint)}{1 + sint} dt =$ $\int_{0}^{\frac{π}{2}} \sin (t) - \sin^{2} (t) dt =$ $\int_{0}^{\frac{π}{2}} \sin (t) dt - \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - \cos (2 t) dt$ $- \cos (t) /_{0}^{\frac{π}{2}} - \frac{1}{2} \cdot t + \frac{1}{4} \cdot \sin (2 t) /_{0}^{\frac{π}{2}} =$ $- (0 - 1) - \frac{1}{2} \cdot \frac{π}{2} + \frac{1}{4} \cdot 0 = 1 - \frac{π}{4}$ $\int_{0}^{1} t \cdot \sqrt{\frac{1 - t}{1 + t}} dt = 1 - \frac{π}{4}$ $A B D U L A Z I Z A B D U V A L I Y E V$
	Answered by ARUNG_Brandon_MBU last updated on 13/Dec/22

	$\int_{0}^{1} t \sqrt{\frac{1 - t}{1 + t}} d t = \int_{0}^{1} t \cdot \frac{1 - t}{\sqrt{1 - t^{2}}} d t$ $= \int_{0}^{1} \frac{t}{\sqrt{1 - t^{2}}} - \int_{0}^{1} \frac{t^{2}}{\sqrt{1 - t^{2}}} d t$ $= - {[\sqrt{1 - t^{2}}]}_{0}^{1} - \int_{0}^{\frac{π}{2}} \sin^{2} θ d θ$ $= 1 - \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 - \cos 2 θ) d θ = 1 - \frac{π}{4}$
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