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Mathematics Question on Integrals of Some Particular Functions

If Δ(x)∣1cos⁡x1−cos⁡x 1+sin⁡xcos⁡x1+sin⁡x−cos⁡x sin⁡xsin⁡x1∣,\Delta (x) \begin{vmatrix}1&\cos x&1 -\cos x\\\ 1+ \sin x& \cos x &1+ \sin x - \cos x\\\ \sin x &\sin x&1\end{vmatrix}, then ∫0π/4Δ(x)dx\int\limits^{\pi / 4}_{0} \Delta\left(x\right)dx is equal to

A

14\frac{1}{4}

B

12\frac{1}{2}

C

0

D

−14 - \frac{1}{4}

Answer

−14 - \frac{1}{4}

Explanation

Solution

Δ(x)=∣1cos⁡x1−cos⁡x 1+sin⁡xcos⁡x1+sin⁡x−cos⁡x sin⁡xsin⁡x1∣,\Delta (x) = \begin{vmatrix}1&\cos x&1 -\cos x\\\ 1+ \sin x & \cos x &1+ \sin x - \cos x\\\ \sin x &\sin x&1\end{vmatrix},
Applying C3→C3+C2−C1C_{3} \to C_{3} + C_{2} - C_{1}
Δ(x)=∣1cos⁡x0 1+sin⁡xcos⁡x0 sin⁡xsin⁡x1∣\Delta\left(x\right) = \begin{vmatrix}1&\cos x &0\\\ 1 +\sin x&\cos x &0\\\ \sin x&\sin x &1\end{vmatrix}
=cos⁡x−cos⁡x(1+sin⁡x)= \cos x - \cos x \left( 1 + \sin x \right)
[∵[\because expanding along C3C_{3}]
=−cos⁡x.sin⁡x=−12sin⁡ 2x= - \cos x. \sin x = - \frac{1}{2} \sin \ 2x
∴∫0π/4Δ(x)dx\therefore \int\limits^{\pi /4}_{0} \Delta\left(x\right)dx
=−12∫0π/4sin⁡ 2x dx= - \frac{1}{2} \int\limits^{\pi/ 4}_{0} \sin \ 2x \ dx
=−12[−cos⁡2x2]0π/4= - \frac{1}{2}\left[- \frac{\cos2x}{2}\right]^{\pi / 4}_{0}
=+12×2[cos⁡π2−cos⁡0∘]= + \frac{1}{2 \times2} \left[ \cos \frac{\pi}{2} - \cos0^{\circ}\right]
=14(0−1)=−14= \frac{1}{4} \left(0-1\right) = - \frac{1}{4}