What is the integral of 7cos^3(4x) ?

The integral of 7cos^3(4x) is 7/4 (sin(4x)-1/3 sin^3(4x))+C

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integral of 7cos^3(4x)

\int 7 cos^{3} (4 x) d x

\frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x)) + C

Solution steps

\int 7 cos^{3} (4 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} (4 x) d x

Apply u-substitution

: \int cos^{3} (u) \frac{1}{4} d u

= 7 \cdot \int cos^{3} (u) \frac{1}{4} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 7 \cdot \frac{1}{4} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= 7 \cdot \frac{1}{4} \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}{4} (\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}{4} (v - \frac{v ^{3}}{3})

Substitute back

= 7 \cdot \frac{1}{4} (sin (4 x) - \frac{sin ^{3} ( 4 x )}{3})

Simplify

7 \cdot \frac{1}{4} (sin (4 x) - \frac{sin ^{3} ( 4 x )}{3}) : \frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x))

= \frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x))

Add a constant to the solution

= \frac{7}{4} (sin (4 x) - \frac{1}{3} sin^{3} (4 x)) + C

t^{2} y^{^{'}} + y = sin (t), y (1) = 0

\frac{\partial}{\partial y} ((1 - x y)^{- 2})

\int \frac{x ^{6}}{x ^{14} - 5} d x

f (x) = \frac{3 x + 5}{1 - x ^{2}}

f (x) = \frac{10 x}{x ^{2} - 6}

What is the integral of 7cos^3(4x) ?
The integral of 7cos^3(4x) is 7/4 (sin(4x)-1/3 sin^3(4x))+C