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

The integral of sin(2x)*cos^2(2x) is -1/6 cos^3(2x)+C

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

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

- \frac{1}{6} cos^{3} (2 x) + C

Solution steps

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

Apply u-substitution

: \int sin (u) cos^{2} (u) \frac{1}{2} d u

= \int sin (u) cos^{2} (u) \frac{1}{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 \int sin (u) cos^{2} (u) d u

Apply u-substitution

: \int - v^{2} d v

= \frac{1}{2} \cdot \int - v^{2} d v

Take the constant out:

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

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

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{2} (- \frac{v ^{3}}{3})

Substitute back

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

Simplify

\frac{1}{2} (- \frac{cos ^{3} ( 2 x )}{3}) : - \frac{1}{6} cos^{3} (2 x)

= - \frac{1}{6} cos^{3} (2 x)

Add a constant to the solution

= - \frac{1}{6} cos^{3} (2 x) + C

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

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

\int \frac{15 x ^{2}}{( 9 - 2 x ^{3} ) ^{6}} d x

\int - e

\int - a

What is the integral of sin(2x)*cos^2(2x) ?
The integral of sin(2x)*cos^2(2x) is -1/6 cos^3(2x)+C