Discussion Forum

If $\triangle (x)=\begin{bmatrix}1 & \cos x&1-\cos x\\1+\sin x & \cos x &1+\sin x-\cos x\\ \sin x&\sin x   &1 \end{bmatrix}$

then   $\int_{0}^{\pi/4} \triangle(x) dx$ is equal to 

Answer:

Explanation:

$\triangle (x)=\begin{bmatrix}1 & \cos x&1-\cos x\\1+\sin x & \cos x &1+\sin x-\cos x\\ \sin x&\sin x   &1 \end{bmatrix}$

Applying  $C_{3} \rightarrow C_{3}+C_{2}-C_{1}$

$\triangle (x)=\begin{bmatrix}1 & \cos x&0\\1+\sin x & \cos x &0\\ \sin x&\sin x   &1 \end{bmatrix}$

$= \cos x-\cos x(1+\sin x)$

   [$\because$  expanding along $C_{3}$ =$- \cos x.\sin x=-\frac{1}{2} \sin 2x$

 $\because$    $\int_{0}^{\pi/4}  \triangle(x) dx=-\frac{1}{2}\int_{0}^{\pi/4} \sin 2x dx$

  $= -\frac{1}{2}\left[\frac{-\cos 2x}{2}\right]_{0}^{\pi/4}$

 $= +\frac{1}{2\times2}\left[ \cos\frac{\pi}{2}-\cos 0^{0}\right]$

 $= \frac{1}{4}(0-1)=-\frac{1}{4}$