∫_0 ^(1/( (√2))) ((x sin^(−1) (x^2 ))/( (√(1−x^4 )))) dx =? ∫2^(−x) tanh (2^(1−x) ) dx =?


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Question Number 115781 by bemath last updated on 28/Sep/20

$\underset{0}{\int^{\frac{1}{\sqrt{2}}}} \frac{x \sin^{- 1} (x^{2})}{\sqrt{1 - x^{4}}} d x = ?$ $\int 2^{- x} \tanh (2^{1 - x}) d x = ?$
	Answered by bobhans last updated on 28/Sep/20
	$I=∫_0 ^(1/( (√2))) ((x.sin^(−1) (x^2 ))/( (√(1−x^4 )))) dx let u = sin^(−1) (x^2 ) → { ((x=(1/( (√2))) →u=(π/6))),((x=0→u=0)) :} I=∫_0 ^(π/6) (1/2)u du = [ (1/4)u^2 ]_0 ^(π/6) = (1/4)×(π^2 /(36)) = (π^2 /(144))$
	$I = \underset{0}{\int^{\frac{1}{\sqrt{2}}}} \frac{x . \sin^{- 1} (x^{2})}{\sqrt{1 - x^{4}}} d x$ $l e t u = \sin^{- 1} (x^{2}) \to {\begin{cases} x = \frac{1}{\sqrt{2}} \to u = \frac{π}{6} \\ x = 0 \to u = 0 \end{cases}$ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{1}{2} u d u = {[\frac{1}{4} u^{2}]}_{0}^{\frac{π}{6}}$ $= \frac{1}{4} \times \frac{π^{2}}{36} = \frac{π^{2}}{144}$
	Answered by Dwaipayan Shikari last updated on 28/Sep/20

	$\int_{0}^{\frac{1}{\sqrt{2}}} \frac{{xsin}^{- 1} (x^{2})}{\sqrt{1 - x^{4}}} dx$ $\frac{1}{2} \int_{0}^{\frac{1}{2}} \frac{\sin^{- 1} (t)}{\sqrt{1 - t^{2}}} dt$ $\frac{1}{4} {[{(\sin^{- 1} (t))}^{2}]}_{0}^{\frac{1}{2}} = \frac{1}{4} . {(\frac{π}{6})}^{2} = \frac{π^{2}}{144}$
	Answered by mathmax by abdo last updated on 28/Sep/20

	$changement \arcsin (x^{2}) = t give x^{2} = sint \Rightarrow x = \sqrt{sint} \Rightarrow$ $\int_{0}^{\frac{1}{\sqrt{2}}} \frac{xarcsin (x^{2})}{\sqrt{1 - x^{4}}} dx = \int_{0}^{\frac{π}{6}} \frac{t \sqrt{sint}}{\sqrt{1 - \sin^{2} t}} \times \frac{cost}{2 \sqrt{sint}} dt$ $= \frac{1}{2} \int_{0}^{\frac{π}{6}} tdt = \frac{1}{2} {[\frac{t^{2}}{2}]}_{0}^{\frac{π}{6}} = \frac{1}{4} \times \frac{π^{2}}{36} = \frac{π^{2}}{144}$
	Answered by Ar Brandon last updated on 28/Sep/20
	$I=∫2^(−x) tanh(2^(1−x) )dx=−(1/(2ln2))∫−2^(1−x) ln2tanh(2^(1−x) )dx =−(1/(2ln2))∫tanh(u)du {(d/dx)(2^(1−x) )=−2^(1−x) ln2} =−(1/(2ln2)){utanh(u)−∫(u/(1−u^2 ))du} =−(1/(2ln2)){utanh(u)+((ln∣1−u^2 ∣)/2)}+C =−(1/(2ln2)){2^(1−x) tanh(2^(1−x) )+((ln∣1−2^(2(1−x)) ∣)/2)}+C$
	$I = \int 2^{- x} \tanh (2^{1 - x}) d x = - \frac{1}{2 \ln 2} \int - 2^{1 - x} \ln 2 \tanh (2^{1 - x}) d x$ $= - \frac{1}{2 \ln 2} \int \tanh (u) du {\frac{d}{d x} (2^{1 - x}) = - 2^{1 - x} \ln 2}$ $= - \frac{1}{2 \ln 2} {utanh (u) - \int \frac{u}{1 - u^{2}} du}$ $= - \frac{1}{2 \ln 2} {utanh (u) + \frac{\ln ∣ 1 - u^{2} ∣}{2}} + C$ $= - \frac{1}{2 \ln 2} {2^{1 - x} \tanh (2^{1 - x}) + \frac{\ln ∣ 1 - 2^{2 (1 - x)} ∣}{2}} + C$
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