Discussion Forum

If $f(x)=\begin{vmatrix}\cos x & 1 & 0 \\1 & 2\cos x & 1 \\0 & 1 & 2\cos x \end{vmatrix}$

$\int_{0}^{\pi/2}f(x)dx$ is equal to

Answer:

Explanation:

f(x)= 4cos³ x - cos x - 2cos x = cos 3x

      (Expansion of determinant)

$\therefore\int_{0}^{\pi/2}f(x)dx =\frac{\sin3x}{3}]^{{\pi/2}}_{0}=-\frac{1}{3}$