Question:
If \(f(x)=\begin{vmatrix} \sin x+\sin 2x+\sin 3x & \sin 2x & \sin 3x \\ 3+4\,\sin x & 3 & 4\,\sin x \\ 1+\sin x & \sin x & 1 \end{vmatrix}\), then \(\int _{0}^{\frac{\pi }{2}}f(x)dx=\)
If \(f(x)=\begin{vmatrix} \sin x+\sin 2x+\sin 3x & \sin 2x & \sin 3x \\ 3+4\,\sin x & 3 & 4\,\sin x \\ 1+\sin x & \sin x & 1 \end{vmatrix}\), then \(\int _{0}^{\frac{\pi }{2}}f(x)dx=\)
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Reduce the determinant with a column operation to get f(x) equal to sin 3x, then integrate.
Updated On: Nov 12, 2026
- \(1\)
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Apply \(C_1\rightarrow C_1-C_2-C_3\) to simplify the determinant.
Step 2: Simplify the determinant
The first column becomes \((\sin x,\ 0,\ 0)\). Expanding along it, \(f(x)=\sin x\,(3-4\sin^{2}x)=3\sin x-4\sin^{3}x=\sin 3x\).
Step 3: Integrate
\(\int_{0}^{\pi/2}\sin 3x\,dx=\left[-\frac{\cos 3x}{3}\right]_{0}^{\pi/2}=\frac{-\cos\frac{3\pi}{2}+\cos 0}{3}=\frac{1}{3}\).
Final Answer:
The value of the integral is \(\frac{1}{3}\). \[ \boxed{\text{(C) }\dfrac{1}{3}} \]
Apply \(C_1\rightarrow C_1-C_2-C_3\) to simplify the determinant.
Step 2: Simplify the determinant
The first column becomes \((\sin x,\ 0,\ 0)\). Expanding along it, \(f(x)=\sin x\,(3-4\sin^{2}x)=3\sin x-4\sin^{3}x=\sin 3x\).
Step 3: Integrate
\(\int_{0}^{\pi/2}\sin 3x\,dx=\left[-\frac{\cos 3x}{3}\right]_{0}^{\pi/2}=\frac{-\cos\frac{3\pi}{2}+\cos 0}{3}=\frac{1}{3}\).
Final Answer:
The value of the integral is \(\frac{1}{3}\). \[ \boxed{\text{(C) }\dfrac{1}{3}} \]
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