Question

If $△=$ $\left|\begin{array}{ccc}2co{s}^{2}x& sin2x& -sinx\\ sin2x& 2si{n}^{2}x& cosx\\ sinx& -cosx& 0\end{array}\right|$
Prove that ${\int }_0^{\pi /2}[△+{△}^{\prime }]dx=\pi$
where ${△}^{\prime }=d/dx\left(△\right).$

Answer 1

Apply $C_1-2sinx{C}_3,C_2+2cosX{C}_3$

$△=$ $\left|\begin{array}{ccc}2& 0& -sinx\\ 0& 2& cosx\\ sinx& -cosx& 0\end{array}\right|$

=$2(co{s}^{2}x+si{n}^x)=2$

$\therefore {△}^{\prime }=0and△+{△}^{\prime }=2$.

$\therefore {\int }_0^{\pi /2}(△+{△}^{\prime })dx={\int }_0^{\pi /2}2dx=[2x{]}_0^{\pi /2}=\pi$.