Question Number 90592 by M±th+et£s last updated on 24/Apr/20
$$\int_{\mathrm{0}} ^{\sqrt{{arccos}\left(\frac{−\mathrm{2}\phi}{\pi}+\mathrm{1}\right)}} {x}\:{sin}\left({x}^{\mathrm{2}} \right)\:{dx} \\ $$
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)](https://www.tinkutara.com/question/Q90593.png)
$$\int_{\mathrm{0}} ^{\sqrt{{arcos}\left(\frac{−\mathrm{2}\varphi}{\pi}+\mathrm{1}\right)}} \:{xsin}\left({x}^{\mathrm{2}} \right){dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left({x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\left(…\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{\mathrm{2}\varphi}{\pi}+\mathrm{1}−\mathrm{1}\right\}\:=\frac{\varphi}{\pi} \\ $$
Commented by M±th+et£s last updated on 25/Apr/20
$${thank}\:{you}\:{sir}\:{nice}\:{work}\: \\ $$
Commented by turbo msup by abdo last updated on 25/Apr/20
$${you}\:{are}\:{welcome} \\ $$