(∫sin^2 πx dx =)? | Mathematics | SolveeIt

$∫sin^2 πx dx =$ ?

$\dfrac{x}{2}-\dfrac{1}{4π} sin2πx+C$

$\dfrac{x}{2}+\dfrac{1}{8π} sin4πx+C$

$\dfrac{x}{8}-\dfrac{1}{4π} cos2πx+C$

$x+\dfrac{1}{2π} sin2πx+C$

$\dfrac{x}{2}-\dfrac{1}{2π} cos2πx+C$

Answer

$\dfrac{x}{2}-\dfrac{1}{4π} sin2πx+C$

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)

Mathematics Question on Integral Calculus