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  4. Let Δ(x)=| cos 2 x cos x sin x - sin x cos x sin x sin 2 x cos x sin x - cos x 0| then ∫ limits0π / 2[Δ(x)+Δ prime(x)] d x equals

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Q. Let $\Delta(x)=\begin{vmatrix}\cos ^2 x & \cos x \sin x & -\sin x \\ \cos x \sin x & \sin ^2 x & \cos x \\ \sin x & -\cos x & 0\end{vmatrix}$ then $\int\limits_0^{\pi / 2}\left[\Delta(x)+\Delta^{\prime}(x)\right] d x$ equals

Determinants

Solution:

Applying $C_1 \rightarrow C_1-\sin x C_3$ and $C_2 \rightarrow C_2+$ $\cos x C_3$, we get
$\Delta(x)=\begin{vmatrix}1 & 0 & -\sin x \\0 & 1 & \cos x \\\sin x & -\cos x & 0
\end{vmatrix}$
Applying $R_3 \rightarrow R_3-\sin x R_1+\cos x R_2$, we get
$\Delta(x) =\begin{vmatrix}1 & 0 & -\sin x \\0 & 1 & \cos x \\0 & 0 & \cos ^2 x+\sin ^2 x\end{vmatrix}=1 $
$\Rightarrow \Delta^{\prime}(x) =0$
Thus, $\int\limits_0^{\pi / 2}\left[\Delta(x)+\Delta^{\prime}(x)\right] d x=\int\limits_0^{\pi / 2} d x=\frac{\pi}{2}$.