Question

If $f\left(x\right)=\left|\begin{array}{ccc}\sin x+\sin 2x+\sin 3x& \sin 2x& \sin 3x\\ 3+4\sin x& 3& 4\sin x\\ 1+\sin x& \sin x& 1\end{array}\right|$, then the value of ${\int }_0^{\pi /2}f\left(x\right)dx$ is

• A. $3$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $0$

Answer 1

The correct option is C $\frac{1}{3}$

We have,
$f\left(x\right)=\left|\begin{array}{ccc}\sin x+\sin 2x+\sin 3x& \sin 2x& \sin 3x\\ 3+4\sin x& 3& 4\sin x\\ 1+\sin x& \sin x& 1\end{array}\right|$

Applying $C_1\to C_1-C_2-C_3$, we get

$f\left(x\right)=\left|\begin{array}{ccc}\sin x+\sin 2x+\sin 3x-\sin 2x-\sin 3x& \sin 2x& \sin 3x\\ 3+4\sin x-3-4\sin x& 3& 4\sin x\\ 1+\sin x-1-\sin x& \sin x& 1\end{array}\right|$

$f\left(x\right)=\left|\begin{array}{ccc}\sin x& \sin 2x& \sin 3x\\ 0& 3& 4\sin x\\ 0& \sin x& 1\end{array}\right|$

$f\left(x\right)=\sin x(3-4{\sin }^{2}x)-\sin 2x(0-0)+\sin 3x(0-0)$

$f\left(x\right)=3\sin x-4{\sin }^{3}x=\sin 3x$

$\therefore I={\int }_0^{\pi /2}f\left(x\right)dx={\int }^{\pi /2}\sin 3xdx$

$I={[-\frac{\cos 3x}{3}]}_0^{\pi /2}$

$I=-\frac{1}{3}\left[\cos \frac{3\pi }{2}-\cos 0\right]=\frac{1}{3}$