Question Number 213139 by Spillover last updated on 31/Oct/24
Answered by MrGaster last updated on 31/Oct/24
![let u=(√x)+(√y)+z⇒dy=(1/(2(√x)))dx+(1/(2(√y)))dy+(1/(2(√z)))dx ⇒0≤u≤((3(√π))/4) ⇒dxdydz=8u^2 du ⇔∫_0 ^((3(√π))/4) sin(u)∙8u^2 du⇒8∫_0 ^((3(√π))/4) u^2 sin(u)du let u=u^2 ,du=2udu,dw=sin(u)du,w=−cos(u) ⇒∫u^2 sin(u)du=−u^2 cos(u)+2∫u cos(u) let v=u,dv=du,dw=cos(u)du,w=sin(u) ⇒∫u cos(u)du=u sin(u)du=u sin(u)+cos(u) ⇒∫u^2 sin(u)du=−u^2 cos(u)+2(u sin(u)+cos(u)) 8−[−(((3(√π))/4))cos(((3(√π))/4))+((3(√π))/2)sin(((3(√π))/4))+2 cos(((3(√π))/4))+2 cos((3(√π))/4)] ⇒8[−((9π)/(16))cos(((3(√π))/4))+((3(√π))/2)sin(((3(√π))/4))+2cos(((3(√π))/4))] ⇔8[(2−((9π)/(16)))cos(((3(√π))/4))+((3(√π))/2)sin(((3(√π))/4))]](https://www.tinkutara.com/question/Q213150.png)
$${let}\:{u}=\sqrt{{x}}+\sqrt{{y}}+{z}\Rightarrow{dy}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}}}{dy}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{z}}}{dx} \\ $$$$\Rightarrow\mathrm{0}\leq{u}\leq\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}} \\ $$$$\Rightarrow{dxdydz}=\mathrm{8}{u}^{\mathrm{2}} {du} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}} \mathrm{sin}\left({u}\right)\centerdot\mathrm{8}{u}^{\mathrm{2}} {du}\Rightarrow\mathrm{8}\int_{\mathrm{0}} ^{\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}} {u}^{\mathrm{2}} \mathrm{sin}\left({u}\right){du} \\ $$$${let}\:{u}={u}^{\mathrm{2}} ,{du}=\mathrm{2}{udu},{dw}=\mathrm{sin}\left({u}\right){du},{w}=−\mathrm{cos}\left({u}\right) \\ $$$$\Rightarrow\int{u}^{\mathrm{2}} \mathrm{sin}\left({u}\right){du}=−{u}^{\mathrm{2}} \mathrm{cos}\left({u}\right)+\mathrm{2}\int{u}\:\mathrm{cos}\left({u}\right) \\ $$$${let}\:{v}={u},{dv}={du},{dw}=\mathrm{cos}\left({u}\right){du},{w}=\mathrm{sin}\left({u}\right) \\ $$$$\Rightarrow\int{u}\:\mathrm{cos}\left({u}\right){du}={u}\:\mathrm{sin}\left({u}\right){du}={u}\:\mathrm{sin}\left({u}\right)+\mathrm{cos}\left({u}\right) \\ $$$$\Rightarrow\int{u}^{\mathrm{2}} \mathrm{sin}\left({u}\right){du}=−{u}^{\mathrm{2}} \mathrm{cos}\left({u}\right)+\mathrm{2}\left({u}\:\mathrm{sin}\left({u}\right)+\mathrm{cos}\left({u}\right)\right) \\ $$$$\mathrm{8}−\left[−\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\mathrm{2}\:\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\mathrm{2}\:\mathrm{cos}\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right] \\ $$$$\Rightarrow\mathrm{8}\left[−\frac{\mathrm{9}\pi}{\mathrm{16}}\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\mathrm{2cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)\right] \\ $$$$\Leftrightarrow\mathrm{8}\left[\left(\mathrm{2}−\frac{\mathrm{9}\pi}{\mathrm{16}}\right)\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)\right] \\ $$
Commented by Frix last updated on 31/Oct/24
$$\mathrm{No}. \\ $$
Answered by Frix last updated on 31/Oct/24
![Using (1) ∫sin (α+(√t)) dt =^([u=(√t)]) 2∫usin (α+u) du =^([by parts.]) =2(sin (α+u) −ucos (α+u))= =2(sin (α+(√t)) −(√t)cos (α+(√t))) (2) Similar ∫cos (α+(√t)) dt=2(cos (α+(√t)) +(√t)sin (α+(√t))) We get 6(4(√π)cos (√π)+(π−4)sin(√π))− −12((√π)cos ((√π)/2) −2sin ((√π)/2))+ +(√π)(π−12)cos ((3(√π))/2) −2(3π−4)sin ((3(√π))/2)](https://www.tinkutara.com/question/Q213154.png)
$$\mathrm{Using} \\ $$$$\left(\mathrm{1}\right) \\ $$$$\int\mathrm{sin}\:\left(\alpha+\sqrt{{t}}\right)\:{dt}\:\overset{\left[{u}=\sqrt{{t}}\right]} {=}\:\mathrm{2}\int{u}\mathrm{sin}\:\left(\alpha+{u}\right)\:{du}\:\overset{\left[\mathrm{by}\:\mathrm{parts}.\right]} {=} \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:\left(\alpha+{u}\right)\:−{u}\mathrm{cos}\:\left(\alpha+{u}\right)\right)= \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:\left(\alpha+\sqrt{{t}}\right)\:−\sqrt{{t}}\mathrm{cos}\:\left(\alpha+\sqrt{{t}}\right)\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Similar} \\ $$$$\int\mathrm{cos}\:\left(\alpha+\sqrt{{t}}\right)\:{dt}=\mathrm{2}\left(\mathrm{cos}\:\left(\alpha+\sqrt{{t}}\right)\:+\sqrt{{t}}\mathrm{sin}\:\left(\alpha+\sqrt{{t}}\right)\right) \\ $$$$\mathrm{We}\:\mathrm{get} \\ $$$$\mathrm{6}\left(\mathrm{4}\sqrt{\pi}\mathrm{cos}\:\sqrt{\pi}+\left(\pi−\mathrm{4}\right)\mathrm{sin}\sqrt{\pi}\right)− \\ $$$$\:\:\:\:\:−\mathrm{12}\left(\sqrt{\pi}\mathrm{cos}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:−\mathrm{2sin}\:\frac{\sqrt{\pi}}{\mathrm{2}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\sqrt{\pi}\left(\pi−\mathrm{12}\right)\mathrm{cos}\:\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\:−\mathrm{2}\left(\mathrm{3}\pi−\mathrm{4}\right)\mathrm{sin}\:\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}} \\ $$