What is the integral of sin^2(x)cos^3(x ?

The integral of sin^2(x)cos^3(x is cos^3(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^3(x

\int sin^{2} (x) cos^{3} (d) x d x

cos^{3} (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^{3} (d) x d x

Take the constant out:

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

= cos^{3} (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^{3} (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^{3} (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^{3} (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

ln^{2} (8 x^{2})

\int \frac{1}{2} ln x^{2} + 1 d x

\int \frac{sin ( x )}{cos ^{2} ( x ) + 1} d x

\int sin (3 x) sin (2 x) sin (x) d x

\frac{1}{s + \frac{1}{a}}

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