What is the integral of 5xsin^2(x) ?

The integral of 5xsin^2(x) is 5(1/2 x(x-1/2 sin(2x))-1/2 ((x^2)/2+1/4 cos(2x)))+C

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integral of 5xsin^2(x)

\int 5 x sin^{2} (x) d x

5 (\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 5 x sin^{2} (x) d x

Take the constant out:

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

= 5 \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

= 5 (\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))

= 5 (\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

= 5 (\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} (sin (2 x) cos (2 y))

\frac{x ^{2}}{1 - x ^{2}}

\int 15 x^{\frac{1}{2}} d x

\int_{- 2}^{2} x^{2} + 3 x - 4 d x

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{7}}

What is the integral of 5xsin^2(x) ?
The integral of 5xsin^2(x) is 5(1/2 x(x-1/2 sin(2x))-1/2 ((x^2)/2+1/4 cos(2x)))+C