What is the integral of 3x*sin^2(x) ?

The integral of 3x*sin^2(x) is 3(1/2 x(x-1/2 sin(2x))-1/2 ((x^2)/2+1/4 cos(2x)))+C

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integral of 3x*sin^2(x)

\int 3 x \cdot sin^{2} (x) d x

3 (\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 3 x sin^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \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

= 3 (\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))

= 3 (\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

= 3 (\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

\frac{\partial}{\partial x} (x y e^{- 1 y})

x^{3}, 4 x

\frac{d}{d x} (x - arccos (x + 1))

y = (7 t - 4), (0, 3)

\frac{\partial}{\partial y} (x \cdot sin (y))

What is the integral of 3x*sin^2(x) ?
The integral of 3x*sin^2(x) is 3(1/2 x(x-1/2 sin(2x))-1/2 ((x^2)/2+1/4 cos(2x)))+C