Question:medium

\[ \int \sin^2x\,dx= \]

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Use \(\sin^2x=\frac{1-\cos2x}{2}\) for integration of \(\sin^2x\).
  • \(\frac{x}{2}+\frac{\sin2x}{4}+c\)
  • \(\frac{x}{2}-\frac{\cos2x}{4}+c\)
  • \(\frac{x}{2}+\frac{\cos2x}{4}+c\)
  • \(\frac{x}{2}-\frac{\sin2x}{4}+c\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
We cannot directly integrate $\sin^2 x$. The standard technique is to use a trigonometric identity to reduce the power of the function to one, making it easily integrable.
Step 2: Key Formula or Approach:
We use the power-reduction formula for sine, which is derived from the double-angle identity for cosine: \[ \cos(2x) = 1 - 2\sin^2 x \] Rearranging this formula to solve for $\sin^2 x$, we get: \[ \sin^2 x = \frac{1 - \cos(2x)}{2} \] Step 3: Detailed Explanation:
The integral is $\int \sin^2 x \,dx$. Substitute the power-reduction formula into the integral: \[ \int \frac{1 - \cos(2x)}{2} \,dx \] Split the integral into two parts: \[ \int \left(\frac{1}{2} - \frac{\cos(2x)}{2}\right) \,dx = \int \frac{1}{2} \,dx - \frac{1}{2} \int \cos(2x) \,dx \] Now integrate each part: \[ \int \frac{1}{2} \,dx = \frac{1}{2}x \] For the second part, $\int \cos(2x) \,dx$, we get $\frac{\sin(2x)}{2}$. So: \[ -\frac{1}{2} \int \cos(2x) \,dx = -\frac{1}{2} \left(\frac{\sin(2x)}{2}\right) = -\frac{\sin(2x)}{4} \] Combine the parts and add the constant of integration, C: \[ \frac{1}{2}x - \frac{\sin(2x)}{4} + C \] Step 4: Final Answer:
The result of the integration is $\frac{x}{2} - \frac{\sin 2x}{4} + c$. Therefore, option (D) is correct.
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