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Evaluate \(\rm \int sin^2 x\;dx\)1. \(\rm \frac{x}{2}+\frac{\sin 2x}{2} + c\)2. \(\rm \frac{x}{2}+\frac{\sin 2x}{4} + c\)3. \(\rm \frac{x}{2}-\frac{\sin 2x}{4} + c\)4. \(\rm \frac{x}{2}+\frac{\cos 2x}{4} + c\) |
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Answer» Correct Answer - Option 3 : \(\rm \frac{x}{2}-\frac{\sin 2x}{4} + c\) Concept: 1 + cos 2x = 2cos2 x 1 - cos 2x = 2sin2 x \(\rm \int \cos x\;dx = \sin x + c\)
Calculation: I = \(\rm \int sin^2 x\;dx\) = \(\rm \int \frac{1-\cos 2x}{2}\;dx\) = \(\rm \frac{1}{2}\int (1-\cos 2x)\;dx\) = \(\rm \frac{1}{2} \left[x-\frac{\sin 2x}{2} \right ] + c\) = \(\rm \frac{x}{2}-\frac{\sin 2x}{4} + c\) |
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