What is the integral of xsin^3(x)cos(x) ?

The integral of xsin^3(x)cos(x) is 1/4 xsin^4(x)-1/4 (-1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x)))+C

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integral of xsin^3(x)cos(x)

\int x sin^{3} (x) cos (x) d x

\frac{1}{4} x sin^{4} (x) - \frac{1}{4} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) + C

Solution steps

\int x sin^{3} (x) cos (x) d x

Apply Integration By Parts

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

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

\int \frac{1}{4} sin^{4} (x) d x = \frac{1}{4} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

= \frac{1}{4} x sin^{4} (x) - \frac{1}{4} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

Add a constant to the solution

= \frac{1}{4} x sin^{4} (x) - \frac{1}{4} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) + C

\frac{d}{d t} (cos (2 t))

\frac{d}{d x} ((x + 1) (2 x + 5)^{4})

\int e^{- \frac{x}{2}} d x

\frac{x ^{2} - 6 x}{x + 1}

y = x^{2} - 2 x, y = x + 4

What is the integral of xsin^3(x)cos(x) ?
The integral of xsin^3(x)cos(x) is 1/4 xsin^4(x)-1/4 (-1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x)))+C