Question

Let $\left|\begin{array}{ccc}2{\cos }^{2}x& \sin \left(2x\right)& -\sin x\\ \sin \left(2x\right)& 2{\sin }^{2}x& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$

Find $\frac{1}{\pi }{\int }_0^{\pi /2}\left[f\right(x)+f{}^{\prime }(x\left)\right]dx$

Answer 1

We have, $f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{2}x& \sin \left(2x\right)& -\sin x\\ \sin \left(2x\right)& 2{\sin }^{2}x& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$

$applying{C}_1\to C_1-2\sin x{C}_{3}and{C}_2\to C_2+2\cos x{C}_{3}wegetf\left(x\right)=\left|\begin{array}{ccc}2& 0& -\sin x\\ 0& 2& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$

$f\left(x\right)=2{\sin }^{2}x+2{\cos }^{2}x=2\Rightarrow f^{\prime }\left(x\right)=0$

$\Rightarrow \frac{1}{\pi }{\int }_0^{\pi /2}\left[f\right(x)+f^{\prime }(x\left)\right]dx=\frac{1}{\pi }{\int }_0^{\pi /2}2dx$

$f\left(x\right)=\frac{1}{\pi }2\left(\frac{\pi }{2}\right)=1$