If $\mathrm{\Delta}(x) = \left| \begin{matrix} 1 & \cos x & 1 - \cos x \ 1 + \sin x & \cos x & 1 + \sin{x - \cos x} \ \sin x & \sin x & 1 \end{matrix} ight|$, then $\int_{0}^{\pi/2}{\mathrm{\Delta}(x)dx}$ is equal to

Question

If $\mathrm{\Delta}(x) = \left| \begin{matrix} 1 & \cos x & 1 - \cos x \ 1 + \sin x & \cos x & 1 + \sin{x - \cos x} \ \sin x & \sin x & 1 \end{matrix} ight|$, then $\int_{0}^{\pi/2}{\mathrm{\Delta}(x)dx}$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Given, $\mathrm{\Delta}(x) = \left| \begin{matrix}
1 & \cos x & 1 - \cos x \
1 + \sin x & \cos x & 1 + \sin{x - \cos x} \
\sin x & \sin x & 1
\end{matrix} ight|$
Applying $C_{3} ightarrow C_{3} + C_{2} - C_{1}$
$= \left| \begin{matrix}
1 & \cos x & 0 \
1 + \sin x & \cos x & 0 \
\sin x & \sin x & 1
\end{matrix} ight|$
$= \cos x - \cos x(1 + \sin x)$
$= - \cos x\ \sin x$
$= - \frac{1}{2}\sin{2x}$
$	herefore\int_{0}^{\pi/2}{\mathrm{\Delta}x\ dx = - \frac{1}{2}}\int_{0}^{\pi/2}{\sin{2x\ dx}}$
$= - \frac{1}{2}{\left\lbrack \  - \frac{\cos{2x}}{2} ightbrack_{0}}^{\pi/2} = - \frac{1}{2}\$

Q. If $\mathrm{\Delta}(x) = \left| \begin{matrix} 1 & \cos x & 1 - \cos x \\ 1 + \sin x & \cos x & 1 + \sin{x - \cos x} \\ \sin x & \sin x & 1 \end{matrix} \right|$, then $\int_{0}^{\pi/2}{\mathrm{\Delta}(x)dx}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?