What is the integral of sin^2(x)x ?

The integral of sin^2(x)x is 1/2 x(x-1/2 sin(2x))-1/2 ((x^2)/2+1/4 cos(2x))+C

integral of sin^2(x)x

\int sin^{2} (x) x d x

\frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)) + C

Solution steps

\int sin^{2} (x) x d x

Apply Integration By Parts

= \frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x

\int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))

= \frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))

Add a constant to the solution

= \frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)) + C

y^{^{'}} = - a y + b

t an g e n t f (x) = 2 x^{2} + x - 1, at x = - 2

\frac{d}{d x} (x e^{(- \frac{x ^{2}}{8})})

d er i v a t i v e f (x) = - \frac{1}{6} (x^{- 6} - x^{12})

x y^{^{'}} + y = x^{\frac{3}{2}} y^{\frac{5}{2}} ln (x)

What is the integral of sin^2(x)x ?
The integral of sin^2(x)x is 1/2 x(x-1/2 sin(2x))-1/2 ((x^2)/2+1/4 cos(2x))+C