Question

If $\Delta \left(x\right)=\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|$, then ${\int }_0^{\pi /2}\Delta \left(x\right)dx$ is equal to

• A. $1/4$
• B. $1/2$
• C. $0$
• D. $-1/2$

Answer 1

The correct option is D $-1/2$

$\Delta \left(x\right)=\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|$

$R_2\to R_2-R_1$

$\Delta \left(x\right)=\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ \sin x& 0& \sin x\\ \sin x& \sin x& 1\end{array}\right|$

$=1(-{\sin }^{2}x)+\cos x({\sin }^{2}x-\sin x)+(1-cosx){\sin }^{2}x$

$=-\cos x\cdot \sin x$

$=-\frac{\sin 2x}{2}$

${\int }_0^{\pi /2}\Delta \left(x\right)dx=-\frac{1}{2}{\int }_0^{\pi /2}\sin 1x\cdot dx$

$=\frac{1}{4}[\cos 2x{]}_0^{\pi /2}=-1/2$