What is the integral of cos^7(2x) ?

The integral of cos^7(2x) is 1/2 (sin(2x)-sin^3(2x)+(3sin^5(2x))/5-(sin^7(2x))/7)+C

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integral of cos^7(2x)

\int cos^{7} (2 x) d x

\frac{1}{2} (sin (2 x) - sin^{3} (2 x) + \frac{3 sin ^{5} ( 2 x )}{5} - \frac{sin ^{7} ( 2 x )}{7}) + C

Solution steps

\int cos^{7} (2 x) d x

Apply u-substitution

: \int cos^{7} (u) \frac{1}{2} d u

= \int cos^{7} (u) \frac{1}{2} 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^{7} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \frac{1}{2} \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}{2} (\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}{2} (v - v^{3} + \frac{3 v ^{5}}{5} - \frac{v ^{7}}{7})

Substitute back

= \frac{1}{2} (sin (2 x) - sin^{3} (2 x) + \frac{3 sin ^{5} ( 2 x )}{5} - \frac{sin ^{7} ( 2 x )}{7})

Add a constant to the solution

= \frac{1}{2} (sin (2 x) - sin^{3} (2 x) + \frac{3 sin ^{5} ( 2 x )}{5} - \frac{sin ^{7} ( 2 x )}{7}) + C

\frac{\partial}{\partial x} (\frac{2 x}{9 - tan ( x )})

\int \frac{x ^{11}}{2 x ^{6} + 2} d x

y^{^{''}} + y^{^{'}} - 6 y = 2 x

t^{2} y^{^{'}} + t y = y

n = 1 \sum \infty \frac{3}{n ^{2} + 3 n}

What is the integral of cos^7(2x) ?
The integral of cos^7(2x) is 1/2 (sin(2x)-sin^3(2x)+(3sin^5(2x))/5-(sin^7(2x))/7)+C