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

The integral of sin^2(bx)cos(bx) is 1/b sin^3(bx)-2/(3b)sin^3(bx)+C

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integral of sin^2(bx)cos(bx)

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

\frac{1}{b} sin^{3} (b x) - \frac{2}{3 b} sin^{3} (b x) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{b} sin^{3} (b x) - \int sin (b x) sin (2 b x) d x

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

\int sin (b x) sin (2 b x) d x = \frac{2}{3 b} sin^{3} (b x)

= \frac{1}{b} sin^{3} (b x) - \frac{2}{3 b} sin^{3} (b x)

Add a constant to the solution

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

\int 3 x - 2 d x

\frac{d}{d x} (\frac{5}{x + 2})

x^{2} - 1, - x + 2, [0, 1]

\int \frac{sin ( 4 e ^{- 3 x} )}{e ^{3 x}} d x

\int \frac{sec ^{2} ( θ )}{1 + 2 tan ( θ )} d θ

What is the integral of sin^2(bx)cos(bx) ?
The integral of sin^2(bx)cos(bx) is 1/b sin^3(bx)-2/(3b)sin^3(bx)+C