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[lilolilo]

$\int_{0}^{\pi ^{2}}\sqrt{x}sin\sqrt{x}dx$

[nucnt772]

$\int_{0}^{\pi ^{2}}\sqrt{x}sin\sqrt{x}dx$

$I=\int_{0}^{\pi ^{2}}\sqrt{x}sin\sqrt{x}dx$

 

đặt: $t=\sqrt{x}$ $\Rightarrow dt=\frac{dx}{2\sqrt{x}}$ $\Rightarrow dx=2tdt$

 

đổi cận:

$x=0\Rightarrow t=0$

$x=\pi ^{2}\Rightarrow t=\pi$

 

$\Rightarrow I=2\int_{0}^{\pi }t^{2}.sintdt$

 

đặt: $\left\{\begin{matrix} &u=t^{2} & \\ &dv=sintdt & \end{matrix}\right.$ $\Rightarrow \left\{\begin{matrix} &du=2tdt & \\ &v=-cost & \end{matrix}\right.$

 

$\Rightarrow I=-2t^{2}.cost|_{0}^{\pi }+4\int_{0}^{\pi }t.costdt$

 

$=2\pi ^{2}+4I_{1}$

 

$I_{1}=\int_{0}^{\pi }t.costdt$

 

đặt: $\left\{\begin{matrix} &u_{1}=t & \\ &dv_{1}=costdt & \end{matrix}\right.$ $\Rightarrow \left\{\begin{matrix} &du_{1}=dt & \\ &v_{1}=sint & \end{matrix}\right.$

 

$\Rightarrow I_{1}=t.sint|_{0}^{\pi }-\int_{0}^{\pi }sintdt$ $=cost|_{0}^{\pi }=-2$

 

$\Rightarrow I=2\pi ^{2}+4.(-2)=2\pi ^{2}-8$

Edited by nucnt772, 08-04-2014 - 22:54.