91影视

Given integral is :

${鈭�}_0^{蟺/2}鈥�{鈭�}_0^{\cos 鈦�y}鈥�{鈭�}_0^{\sin 鈦�y}鈥�(2x+y)dzdxdy$

We have to evaluate it.

Given

${鈭�}_0^{蟺/2}鈥�{鈭�}_0^{\cos 鈦�y}鈥�{鈭�}_0^{\sin 鈦�y}鈥�(2x+y)dzdxdy$

$\begin{array}{r}={鈭�}_0^{\frac{蟺}{2}}鈥�{鈭�}_0^{\cos 鈦�y}鈥�\left[{鈭�}_0^{\sin 鈦�y}鈥�(2x+y)dz\right]dxdy\\ ={鈭�}_0^{\frac{蟺}{2}}鈥�{鈭�}_0^{\cos 鈦�y}鈥�\left[2x{鈭�}_0^{\sin 鈦�y}鈥�dz+y{鈭�}_0^{\sin 鈦�y}鈥�dz\right]dxdy\\ ={鈭�}_0^{\frac{蟺}{2}}鈥�{鈭�}_0^{\cos 鈦�y}鈥�\left[2x(z{)}_0^{\sin 鈦�y}+y(z{)}_0^{\sin 鈦�y}\right]dxdy\\ ={鈭�}_0^{\frac{蟺}{2}}鈥�{鈭�}_0^{\cos 鈦�y}鈥�[2x\sin 鈦�y+y\sin 鈦�y]dxdy\end{array}$

Integrate with respect to x

$\begin{array}{r}={鈭�}_0^{\frac{蟺}{2}}鈥�\left[{鈭�}_0^{\cos 鈦�y}鈥�2x\sin 鈦�ydx+{鈭�}_0^{\cos 鈦�y}鈥�y\sin 鈦�ydx\right]dy\\ ={鈭�}_0^{\frac{蟺}{2}}鈥�\left[2\sin 鈦�y{鈭�}_0^{\cos 鈦�y}鈥�xdx+y\sin 鈦�y{鈭�}_0^{\cos 鈦�y}鈥�dx\right]dy\\ ={鈭�}_0^{\frac{蟺}{2}}鈥�\left[2\sin 鈦�y{\left(\frac{x^2}{2}\right)}_0^{\cos 鈦�y}+y\sin 鈦�y(x{)}_0^{\cos 鈦�y}\right]dy\\ ={鈭�}_0^{\frac{蟺}{2}}鈥�\left[2\sin 鈦�y\left(\frac{{\cos }^2鈦�y}{2}\right)+y\sin 鈦�y(\cos 鈦�y)\right]dy\end{array}$

Integrate with respect to y

$\begin{array}{r}={鈭�}_0^{\frac{蟺}{2}}鈥�\left(\sin 鈦�y{\cos }^2鈦�y+y\sin 鈦�y\cos 鈦�y\right)dy\\ ={鈭�}_0^{\frac{蟺}{2}}鈥�\sin 鈦�y{\cos }^2鈦�y+\frac{1}{2}{鈭�}_0^{\frac{蟺}{2}}鈥�y\sin 鈦�2ydy\\ ={\left(\frac{鈭�{\cos }^3鈦�y}{3}\right)}_0^{\frac{蟺}{2}}+\frac{1}{2}\left[{\left(\frac{鈭�y\cos 鈦�2y}{2}+\frac{\sin 鈦�2y}{4}\right)}_0^{\frac{蟺}{2}}\right]\\ =\frac{鈭�1}{3}(0鈭�1)+\frac{1}{2}\left[\frac{鈭�1}{2}\left(\frac{蟺}{2}\cos 鈦�蟺鈭�0\right)+\frac{1}{4}(\sin 鈦�蟺鈭�\cos 鈦�0)\right]\\ =\frac{1}{3}+\frac{1}{2}\left[\frac{蟺}{4}\right]\\ =\frac{蟺}{8}+\frac{1}{3}\end{array}$