Question

If we are given a determinant as $\Delta =\left| \begin{matrix}
   2{{\cos }^{2}}x & \sin 2x & -\sin x \\
   \sin 2x & 2{{\sin }^{2}}x & \cos x \\
   \sin x & -\cos x & 0 \\
\end{matrix} \right|$, prove that $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \Delta +{{\Delta }^{'}} \right]}dx=\pi $ where ${{\Delta }^{'}}=\dfrac{d\left( \Delta \right)}{dx}$.

Answer 1

Hint: We find the determinant value of the given form. Then we need to find the differentiation of the determinant. We put the values in the integration of the equation to prove the result. We find the value without integrating like the matrix form.

Complete step-by-step solution:
We have been given the determinant $\Delta =\left| \begin{matrix}
   2{{\cos }^{2}}x & \sin 2x & -\sin x \\
   \sin 2x & 2{{\sin }^{2}}x & \cos x \\
   \sin x & -\cos x & 0 \\
\end{matrix} \right|$. We find the determinant value. We take the first column and multiply every element of that column with its cofactor. We find cofactor of a particular element by omitting that row and column of that particular element. We take the rest of the determinant.
 $\Delta =2{{\cos }^{2}}x\left( {{\cos }^{2}}x \right)+\sin 2x\left( \sin x\cos x \right)+\sin x\left( \sin 2x\cos x+2{{\sin }^{3}}x \right)$.
From the standard trigonometric identities, we know that
 \[\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)=1\].
We use different trigonometric formulas and various simplifications to find the value of the determinant. First, we take $2\sin x$ common and use the formula $\sin 2x=2\sin x\cos x$.
\[\begin{align}
  & \Delta =2{{\cos }^{2}}x\left( {{\cos }^{2}}x \right)+\sin 2x\left( \sin x\cos x \right)+\sin x\left( \sin 2x\cos x+2{{\sin }^{3}}x \right) \\
 & \Rightarrow \Delta =2{{\cos }^{2}}x\left( {{\cos }^{2}}x \right)+2\sin x\cos x\left( \sin x\cos x \right)+2{{\sin }^{2}}x\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right) \\
\end{align}\]
Then we take $2{{\cos }^{2}}x$ common out of the first two terms.
\[\begin{align}
  & \Delta =2{{\cos }^{2}}x\left( {{\cos }^{2}}x \right)+2\sin x\cos x\left( \sin x\cos x \right)+2{{\sin }^{2}}x\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right) \\
 & \Rightarrow \Delta =2{{\cos }^{2}}x\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)+2{{\sin }^{2}}x \\
\end{align}\]
We use the formula of \[\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)=1\] and get
\[\begin{align}
  & \Delta =2{{\cos }^{2}}x\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)+2{{\sin }^{2}}x \\
 & \Rightarrow \Delta =2{{\cos }^{2}}x+2{{\sin }^{2}}x \\
 & \Rightarrow \Delta =2\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)=2 \\
\end{align}\]
We have been asked to prove $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \Delta +{{\Delta }^{'}} \right]}dx=\pi $. The determinant value is constant, which means that differentiation of that determinant will be 0. So, it implies that ${{\Delta }^{'}}=\dfrac{d\left( \Delta \right)}{dx}=\dfrac{d\left( 2 \right)}{dx}=0$.
Now we replace the values of the determinant and its differentiation in the integration
$\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \Delta +{{\Delta }^{'}} \right]}dx$.
After replacing the values, we evaluate the value of the integration.
\[\begin{align}
  & \int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \Delta +{{\Delta }^{'}} \right]}dx \\
 & =\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 2+0 \right]}dx \\
 & =\left[ 2x \right]_{0}^{\dfrac{\pi }{2}} \\
 & =2\times \dfrac{\pi }{2} \\
 & =\pi \\
\end{align}\].
Thus proved.

Note: We don’t need to apply integration in the determinant itself. It gives value. In the case of matrices, we need to use matrix differentiation or integration. In our case, we just convert it to its simplest form to find the solution. The matrix integration will work for column-wise integration of three columns one by one keeping others intact.