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Question Number 90592 by M±th+et£s last updated on 24/Apr/20

∫_0 ^(√(arccos(((−2φ)/π)+1))) x sin(x^2 ) dx

$$\int_{\mathrm{0}} ^{\sqrt{{arccos}\left(\frac{−\mathrm{2}\phi}{\pi}+\mathrm{1}\right)}} {x}\:{sin}\left({x}^{\mathrm{2}} \right)\:{dx} \\ $$

Commented by mathmax by abdo last updated on 24/Apr/20

∫_0 ^(√(arcos(((−2ϕ)/π)+1)))  xsin(x^2 )dx =[−(1/2)cos(x^2 )]_0 ^((...))   =−(1/2){−((2ϕ)/π)+1−1} =(ϕ/π)

$$\int_{\mathrm{0}} ^{\sqrt{{arcos}\left(\frac{−\mathrm{2}\varphi}{\pi}+\mathrm{1}\right)}} \:{xsin}\left({x}^{\mathrm{2}} \right){dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left({x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\left(...\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{\mathrm{2}\varphi}{\pi}+\mathrm{1}−\mathrm{1}\right\}\:=\frac{\varphi}{\pi} \\ $$

Commented by M±th+et£s last updated on 25/Apr/20

thank you sir nice work

$${thank}\:{you}\:{sir}\:{nice}\:{work}\: \\ $$

Commented by turbo msup by abdo last updated on 25/Apr/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

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