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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2} cos^{3} (u) d u

= \int \frac{1}{2} cos^{3} (u) 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 cos^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int 1 - v^{2} d v

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

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

Substitute back

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (4 + ln (x y - 5))

\int 4^{k} d k

f (θ) = \frac{sin ( θ )}{1 + cos ( θ )}

\frac{d}{d x} (2 x^{3} + 15 x^{2} - 84 x + 10)

30 e^{x^{2}} x

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