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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int - v^{2} d v

= 2 \cdot \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

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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