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

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

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integral of xsin^2(x)

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

= \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

\int (x^{2} - 1) e^{2 x} d x

y^{^{''}} - 6 y^{^{'}} + 4 y = 0

\int_{0}^{x} e^{t^{2}} d t

\int (4 + 3 x) e^{- x} d x

\frac{\partial}{\partial x} (2 x e^{x})

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