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

The integral of xsin^2(x)cos^2(x) is 1/8 x(x-1/4 sin(4x))-1/8 ((x^2)/2+1/16 cos(4x))+C

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integral of xsin^2(x)cos^2(x)

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

\frac{1}{8} x (x - \frac{1}{4} sin (4 x)) - \frac{1}{8} (\frac{x ^{2}}{2} + \frac{1}{16} cos (4 x)) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{8} x (x - \frac{1}{4} sin (4 x)) - \int \frac{1}{8} (x - \frac{1}{4} sin (4 x)) d x

= \frac{1}{8} x (x - \frac{1}{4} sin (4 x)) - \int \frac{1}{8} (x - \frac{1}{4} sin (4 x)) d x

\int \frac{1}{8} (x - \frac{1}{4} sin (4 x)) d x = \frac{1}{8} (\frac{x ^{2}}{2} + \frac{1}{16} cos (4 x))

= \frac{1}{8} x (x - \frac{1}{4} sin (4 x)) - \frac{1}{8} (\frac{x ^{2}}{2} + \frac{1}{16} cos (4 x))

Add a constant to the solution

= \frac{1}{8} x (x - \frac{1}{4} sin (4 x)) - \frac{1}{8} (\frac{x ^{2}}{2} + \frac{1}{16} cos (4 x)) + C

\frac{d y}{d x} = \frac{( x y + 7 x - y - 7 )}{x y - 2 x + 8 y - 16}

\int 8 \frac{sin ( 2 x )}{sin ( x )} d x

\int_{0}^{1} (4 x - 4 x) d x

\frac{d y}{d x} + 12 y = 8

\frac{d}{d x} (\frac{sin ( x ^{2} )}{ln ( 2 x + 8 )})

What is the integral of xsin^2(x)cos^2(x) ?
The integral of xsin^2(x)cos^2(x) is 1/8 x(x-1/4 sin(4x))-1/8 ((x^2)/2+1/16 cos(4x))+C