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

\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

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