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

The integral of sin^3(x)cos(x is cos(d)(-xcos(x)+1/3 xcos^3(x)+2/3 sin(x)+1/9 sin^3(x))+C

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integral of sin^3(x)cos(x

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

cos (d) (- x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x)) + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: x (- cos (x) + \frac{1}{3} cos^{3} (x)) - \int - cos (x) + \frac{1}{3} cos^{3} (x) d x

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

\int - cos (x) + \frac{1}{3} cos^{3} (x) d x = - \frac{2}{3} sin (x) - \frac{1}{9} sin^{3} (x)

= cos (d) (x (- cos (x) + \frac{1}{3} cos^{3} (x)) - (- \frac{2}{3} sin (x) - \frac{1}{9} sin^{3} (x)))

Expand

x (- cos (x) + \frac{1}{3} cos^{3} (x)) - (- \frac{2}{3} sin (x) - \frac{1}{9} sin^{3} (x)) : - x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x)

= cos (d) (- x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x))

Add a constant to the solution

= cos (d) (- x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x)) + C

x^{2} - 2 x + 4

\int (7 x^{3} + 4 x^{2}) d x

f (x) = 4 x - x^{2}

\int 36 sin^{2} (x) d x

(1 + t) (4 t^{2} - 9)

What is the integral of sin^3(x)cos(x ?
The integral of sin^3(x)cos(x is cos(d)(-xcos(x)+1/3 xcos^3(x)+2/3 sin(x)+1/9 sin^3(x))+C