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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u}{2} d u

= \frac{1}{2} \cdot \int \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

u = 1 - cos (2 x)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{( 1 - cos ( 2 x ) ) ^{2}}{2}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{( 1 - cos ( 2 x ) ) ^{2}}{2} : \frac{1}{2} sin^{4} (x)

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

Add a constant to the solution

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

\int 9 sec^{- 2} (x t a) n^{3} x d x

\int \frac{x ^{2} - 9}{x ^{4}} d x

\int x - 1 - ln (x) d x

\frac{\partial}{\partial y} (y + 3)

y = x^{2} - x, y = 2 (x + 2)

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