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

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

$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\:\frac{{x}\:\mathrm{sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:{dx}\:=? \\ $$$$\int\mathrm{2}^{−{x}} \:\mathrm{tanh}\:\left(\mathrm{2}^{\mathrm{1}−{x}} \right)\:{dx}\:=? \\ $$

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))

$${I}=\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\:\frac{{x}.\mathrm{sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:{dx} \\ $$$${let}\:{u}\:=\:\mathrm{sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)\:\rightarrow\begin{cases}{{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\rightarrow{u}=\frac{\pi}{\mathrm{6}}}\\{{x}=\mathrm{0}\rightarrow{u}=\mathrm{0}}\end{cases} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{6}}} {\int}}\:\frac{\mathrm{1}}{\mathrm{2}}{u}\:{du}\:=\:\left[\:\frac{\mathrm{1}}{\mathrm{4}}{u}^{\mathrm{2}} \:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{144}} \\ $$

Answered by Dwaipayan Shikari last updated on 28/Sep/20

∫_0 ^(1/( (√2))) ((xsin^(−1) (x^2 ))/( (√(1−x^4 ))))dx (1/2)∫_0 ^(1/2) ((sin^(−1) (t))/( (√(1−t^2 ))))dt (1/4)[(sin^(−1) (t))^2 ]_0 ^(1/2) =(1/4).((π/6))^2 =(π^2 /(144))

$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \frac{\mathrm{xsin}^{−\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\mathrm{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{t}\right)}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{t}\right)\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{4}}.\left(\frac{\pi}{\mathrm{6}}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{144}} \\ $$

Answered by mathmax by abdo last updated on 28/Sep/20

changement arcsin(x^2 )=t give x^2 =sint ⇒x =(√(sint)) ⇒ ∫_0 ^(1/(√2)) ((xarcsin(x^2 ))/(√(1−x^4 )))dx =∫_0 ^(π/6) ((t(√(sint)))/(√(1−sin^2 t)))×((cost)/(2(√(sint)))) dt =(1/2) ∫_0 ^(π/6) tdt =(1/2)[(t^2 /2)]_0 ^(π/6) =(1/4)×(π^2 /(36)) =(π^2 /(144))

$$\mathrm{changement}\:\mathrm{arcsin}\left(\mathrm{x}^{\mathrm{2}} \right)=\mathrm{t}\:\:\mathrm{give}\:\mathrm{x}^{\mathrm{2}} \:=\mathrm{sint}\:\Rightarrow\mathrm{x}\:=\sqrt{\mathrm{sint}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\frac{\mathrm{xarcsin}\left(\mathrm{x}^{\mathrm{2}} \right)}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{\mathrm{t}\sqrt{\mathrm{sint}}}{\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}}×\frac{\mathrm{cost}}{\mathrm{2}\sqrt{\mathrm{sint}}}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\mathrm{tdt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{144}} \\ $$

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

$$\mathcal{I}=\int\mathrm{2}^{−{x}} \mathrm{tanh}\left(\mathrm{2}^{\mathrm{1}−{x}} \right)\mathrm{d}{x}=−\frac{\mathrm{1}}{\mathrm{2ln2}}\int−\mathrm{2}^{\mathrm{1}−{x}} \mathrm{ln2tanh}\left(\mathrm{2}^{\mathrm{1}−{x}} \right)\mathrm{d}{x} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2ln2}}\int\mathrm{tanh}\left(\mathrm{u}\right)\mathrm{du}\:\:\:\:\:\:\:\:\:\:\:\left\{\frac{\mathrm{d}}{\mathrm{d}{x}}\left(\mathrm{2}^{\mathrm{1}−{x}} \right)=−\mathrm{2}^{\mathrm{1}−{x}} \mathrm{ln2}\right\} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2ln2}}\left\{\mathrm{utanh}\left(\mathrm{u}\right)−\int\frac{\mathrm{u}}{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }\mathrm{du}\right\} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2ln2}}\left\{\mathrm{utanh}\left(\mathrm{u}\right)+\frac{\mathrm{ln}\mid\mathrm{1}−\mathrm{u}^{\mathrm{2}} \mid}{\mathrm{2}}\right\}+\mathcal{C} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2ln2}}\left\{\mathrm{2}^{\mathrm{1}−{x}} \mathrm{tanh}\left(\mathrm{2}^{\mathrm{1}−{x}} \right)+\frac{\mathrm{ln}\mid\mathrm{1}−\mathrm{2}^{\mathrm{2}\left(\mathrm{1}−{x}\right)} \mid}{\mathrm{2}}\right\}+\mathcal{C} \\ $$

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