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

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

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integral of 2xsin^2(x)

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

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

Solution steps

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

Take the constant out:

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

= 2 \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

= 2 (\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))

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = - x

x^{2} \frac{d y}{d x} - 2 x y = 3 y^{4}

y = (\frac{x ^{2} + 6}{x ^{2} - 6})^{3}

\frac{d}{d x} (arcsin (3 x) + arccos (3 x))

\frac{\partial}{\partial x} (sin (θ))

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