Question:
\(∫sin^2 πx dx =\)?
\(∫sin^2 πx dx =\)?
Updated On: Aug 1, 2024
\(\dfrac{x}{2}-\dfrac{1}{4π} sin2πx+C\)
\(\dfrac{x}{2}+\dfrac{1}{8π} sin4πx+C\)
\(\dfrac{x}{8}-\dfrac{1}{4π} cos2πx+C\)
\(x+\dfrac{1}{2π} sin2πx+C\)
\(\dfrac{x}{2}-\dfrac{1}{2π} cos2πx+C\)
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The Correct Option is A
Solution and Explanation
\(∫sin^2 πx dx \)
\(=∫(\dfrac{1}{2}-\dfrac{1}{2} Cos(2πx)) dx\)
\(=∫\dfrac{1}{2}dx-∫Cos(2πx)dx \)
\(=\dfrac{x}{2} -\dfrac{Sin2πx}{4π} +C \) (Ans)
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