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Question: If a determinant is given as \(\Delta (x) = \left| {\begin{array}{*{20}{c}} 1&{\cos x}&{1 - \cos...

If a determinant is given as \Delta (x) = \left| {\begin{array}{*{20}{c}} 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 ∫0π/4Δ(x)dx\int_0^{\pi /4} {\Delta (x)dx} is equal to
(A). 14\dfrac{1}{4}
(B). 12\dfrac{1}{2}
(C). 00
(D). −14 - \dfrac{1}{4}

Explanation

Solution

Hint- In this question, firstly we have to find the value of the given determinant. Simplify the determinant by applying column operations and solve the determinant and then apply the formula for definite Integration of trigonometric functions to get the answer.

Complete step-by-step solution -
Given, \Delta (x) = \left| {\begin{array}{*{20}{c}} 1&{\cos x}&{1 - \cos x} \\\ {1 + \sin x}&{\cos x}&{1 + \sin x - \cos x} \\\ {\sin x}&{\sin x}&1 \end{array}} \right|
Use column operation C1→C1−C2{C_1} \to {C_1} - {C_2}
\Rightarrow \Delta (x) = \left| {\begin{array}{*{20}{c}} {1 - \cos x}&{\cos x}&{1 - \cos x} \\\ {1 + \sin x - \cos x}&{\cos x}&{1 + \sin x - \cos x} \\\ 0&{\sin x}&1 \end{array}} \right|
Now, use column operation C1→C1−C3{C_1} \to {C_1} - {C_3}
\Rightarrow \Delta (x) = \left| {\begin{array}{*{20}{c}} 0&{\cos x}&{1 - \cos x} \\\ 0&{\cos x}&{1 + \sin x - \cos x} \\\ { - 1}&{\sin x}&1 \end{array}} \right|
Now solving the determinant with respect to the column 1 we get,
⇒Δ(x)=0−0−1×[cos⁡x(1+sin⁡x−cos⁡x)−cos⁡x(1−cos⁡x)]\Rightarrow \Delta (x) = 0 - 0 - 1 \times \left[ {\cos x(1 + \sin x - \cos x) - \cos x(1 - \cos x)} \right]
⇒Δ(x)=−1×[cos⁡x+cos⁡xsin⁡x−cos⁡2x−cos⁡x+cos⁡2x]\Rightarrow \Delta (x) = - 1 \times \left[ {\cos x + \cos x\sin x - {{\cos }^2}x - \cos x + {{\cos }^2}x} \right]
⇒Δ(x)=−1×[cos⁡xsin⁡x]\Rightarrow \Delta (x) = - 1 \times \left[ {\cos x\sin x} \right]
⇒Δ(x)=−cos⁡xsin⁡x\Rightarrow \Delta (x) = - \cos x\sin x
⇒Δ(x)=−2cos⁡xsin⁡x2\Rightarrow \Delta (x) = - \dfrac{{2\cos x\sin x}}{2}
⇒Δ(x)=−sin⁡2x2\Rightarrow \Delta (x) = - \dfrac{{\sin 2x}}{2}
Integrate both sides with respect to x over the definite integral where x varies from 0 to π4\dfrac{\pi }{4}
⇒∫0π/4Δ(x)=∫0π/4−12(sin⁡(2x))dx\Rightarrow \int_0^{\pi /4} {\Delta (x)} = \int_0^{\pi /4} {\dfrac{{ - 1}}{2}\left( {\sin (2x)} \right)dx}
=−12∫0π/4sin⁡(2x)dx= \dfrac{{ - 1}}{2}\int_0^{\pi /4} {\sin (2x)dx}
We know that ∫cdcos⁡(ax)dx=[−sin⁡(ax)a]cd=−1a[sin⁡(ad)−sin⁡(ac)]\int_c^d {\cos (ax)dx = } \left[ {\dfrac{{ - \sin (ax)}}{a}} \right]_c^d = - \dfrac{1}{a}\left[ {\sin (ad) - \sin (ac)} \right]
=−12[−cos⁡2x2]0π/4= \dfrac{{ - 1}}{2}\left[ { - \dfrac{{\cos 2x}}{2}} \right]_0^{\pi /4}
=14[cos⁡2x]0π/4= \dfrac{1}{4}\left[ {\cos 2x} \right]_0^{\pi /4}
=14[cos⁡2π4−cos⁡0]= \dfrac{1}{4}\left[ {\cos \dfrac{{2\pi }}{4} - \cos 0} \right]
We know that cos⁡0=1\cos 0 = 1
=14[cos⁡π2−1]= \dfrac{1}{4}\left[ {\cos \dfrac{\pi }{2} - 1} \right]
We know that cos⁡π2=1\cos \dfrac{\pi }{2} = 1
=14[0−1]= \dfrac{1}{4}\left[ {0 - 1} \right]
=−14= - \dfrac{1}{4}
Hence, ∫0π/4Δ(x)dx=−14\int_0^{\pi /4} {\Delta (x)dx} = - \dfrac{1}{4}
∴\therefore Option D. −14 - \dfrac{1}{4} is the correct answer.

Note- For these types of questions, one has to remember all the properties of determinant and integration to proceed. It is better to use columns/rows operations for simplification. Moreover, one must be knowing that ∫cdcos⁡(ax)dx=[−sin⁡(ax)a]cd=−1a[sin⁡(ad)−sin⁡(ac)]\int_c^d {\cos (ax)dx = } \left[ {\dfrac{{ - \sin (ax)}}{a}} \right]_c^d = - \dfrac{1}{a}\left[ {\sin (ad) - \sin (ac)} \right] , cos⁡0=1\cos 0 = 1 , cos⁡π2=1\cos \dfrac{\pi }{2} = 1 and how to solve the determinant. In these types of questions students start to solve given determinants directly which makes the solution complicated, so this is needed to be careful here.