What is the integral of sin^2(x)cos^4(x ?

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

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integral of sin^2(x)cos^4(x

\int sin^{2} (x) cos^{4} (d) x d x

cos^{4} (d) (\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 sin^{2} (x) cos^{4} (d) x d x

Take the constant out:

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

= cos^{4} (d) \cdot \int sin^{2} (x) 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

= cos^{4} (d) (\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))

= cos^{4} (d) (\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

= cos^{4} (d) (\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{d}{d x} ((x - 4) (x + 3))

\frac{d}{d x} (\frac{x + 3}{2 x + 1})

3.8 t e^{2.4 t}

\frac{\partial}{\partial x} (\frac{- k y}{x ^{2} + y ^{2}})

5 - 3 t

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