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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

= \frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

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

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

\int (9 - 5 e^{x} + \frac{3}{x ^{2}}) d x

x \to - 1 lim (\frac{x - 2}{x + 1})

\frac{1}{z ^{4}}

\frac{d y}{d t} + e^{t} y = 16 e^{t}

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