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

The integral of cos^7(4x) is 1/4 (sin(4x)-sin^3(4x)+(3sin^5(4x))/5-(sin^7(4x))/7)+C

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integral of cos^7(4x)

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

\frac{1}{4} (sin (4 x) - sin^{3} (4 x) + \frac{3 sin ^{5} ( 4 x )}{5} - \frac{sin ^{7} ( 4 x )}{7}) + C

Solution steps

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

Apply u-substitution

: \int cos^{7} (u) \frac{1}{4} d u

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{4} \cdot \int (1 - sin^{2} (u))^{3} cos (u) d u

Apply u-substitution

: \int (1 - v^{2})^{3} d v

= \frac{1}{4} \cdot \int (1 - v^{2})^{3} d v

Expand

(1 - v^{2})^{3} : 1 - 3 v^{2} + 3 v^{4} - v^{6}

= \frac{1}{4} \cdot \int 1 - 3 v^{2} + 3 v^{4} - v^{6} 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}{4} (\int 1 d v - \int 3 v^{2} d v + \int 3 v^{4} d v - \int v^{6} d v)

\int 1 d v = v

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

\int 3 v^{4} d v = \frac{3 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{4} (v - v^{3} + \frac{3 v ^{5}}{5} - \frac{v ^{7}}{7})

Substitute back

= \frac{1}{4} (sin (4 x) - sin^{3} (4 x) + \frac{3 sin ^{5} ( 4 x )}{5} - \frac{sin ^{7} ( 4 x )}{7})

Add a constant to the solution

= \frac{1}{4} (sin (4 x) - sin^{3} (4 x) + \frac{3 sin ^{5} ( 4 x )}{5} - \frac{sin ^{7} ( 4 x )}{7}) + C

\frac{\partial}{\partial x} (4 x^{4} y)

\int 2 x \cdot cos (x) d x

(3, 6), (5, 5)

\frac{d ^{2}}{d x ^{2}} (\frac{6 x ^{5}}{5} - 3 x)

y^{^{''}} - 6 y^{^{'}} + 10 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

What is the integral of cos^7(4x) ?
The integral of cos^7(4x) is 1/4 (sin(4x)-sin^3(4x)+(3sin^5(4x))/5-(sin^7(4x))/7)+C