What is the integral of 7cos^3(3x) ?

The integral of 7cos^3(3x) is 7/3 (sin(3x)-1/3 sin^3(3x))+C

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integral of 7cos^3(3x)

\int 7 cos^{3} (3 x) d x

\frac{7}{3} (sin (3 x) - \frac{1}{3} sin^{3} (3 x)) + C

Solution steps

\int 7 cos^{3} (3 x) d x

Take the constant out:

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

= 7 \cdot \int cos^{3} (3 x) d x

Apply u-substitution

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

= 7 \cdot \int cos^{3} (u) \frac{1}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 7 \cdot \frac{1}{3} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 7 \cdot \frac{1}{3} \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

= 7 \cdot \frac{1}{3} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

= 7 \cdot \frac{1}{3} (v - \frac{v ^{3}}{3})

Substitute back

= 7 \cdot \frac{1}{3} (sin (3 x) - \frac{sin ^{3} ( 3 x )}{3})

Simplify

7 \cdot \frac{1}{3} (sin (3 x) - \frac{sin ^{3} ( 3 x )}{3}) : \frac{7}{3} (sin (3 x) - \frac{1}{3} sin^{3} (3 x))

= \frac{7}{3} (sin (3 x) - \frac{1}{3} sin^{3} (3 x))

Add a constant to the solution

= \frac{7}{3} (sin (3 x) - \frac{1}{3} sin^{3} (3 x)) + C

x y^{^{'}} + 3 y = 0

x \to 0 + lim (3)

f (x) = cos (2 sin (x))

y = ln (e^{x} + x e^{x})

\frac{d}{d x} (3 (9 x - 4)^{4})

What is the integral of 7cos^3(3x) ?
The integral of 7cos^3(3x) is 7/3 (sin(3x)-1/3 sin^3(3x))+C