Question:
\[
\int \sin^2x\,dx=
\]
\[
\int \sin^2x\,dx=
\]
Show Hint
Use \(\sin^2x=\frac{1-\cos2x}{2}\) for integration of \(\sin^2x\).
Updated On: May 7, 2026
- \(\frac{x}{2}+\frac{\sin2x}{4}+c\)
- \(\frac{x}{2}-\frac{\cos2x}{4}+c\)
- \(\frac{x}{2}+\frac{\cos2x}{4}+c\)
- \(\frac{x}{2}-\frac{\sin2x}{4}+c\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use identity: \[ \sin^2x=\frac{1-\cos2x}{2} \]
Step 2: \[ \int\sin^2x\,dx = \int\frac{1-\cos2x}{2}\,dx \]
Step 3: \[ =\frac{1}{2}\int 1\,dx-\frac{1}{2}\int\cos2x\,dx \]
Step 4: \[ =\frac{x}{2}-\frac{1}{2}\cdot \frac{\sin2x}{2}+c \] \[ =\frac{x}{2}-\frac{\sin2x}{4}+c \] \[ \boxed{\frac{x}{2}-\frac{\sin2x}{4}+c} \]
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