What is the integral of (cos^5(6/x))/(x^2) ?

The integral of (cos^5(6/x))/(x^2) is -1/6 (sin(6/x)-(2sin^3(6/x))/3+(sin^5(6/x))/5)+C

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integral of (cos^5(6/x))/(x^2)

\int \frac{cos ^{5} ( \frac{6}{x} )}{x ^{2}} d x

- \frac{1}{6} (sin (\frac{6}{x}) - \frac{2 sin ^{3} ( \frac{6}{x} )}{3} + \frac{sin ^{5} ( \frac{6}{x} )}{5}) + C

Solution steps

\int \frac{cos ^{5} ( \frac{6}{x} )}{x ^{2}} d x

Apply u-substitution

: \int - \frac{cos ^{5} ( u )}{6} d u

= \int - \frac{cos ^{5} ( u )}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int cos^{5} (u) d u

Rewrite using trig identities

= - \frac{1}{6} \cdot \int (1 - sin^{2} (u))^{2} cos (u) d u

Apply u-substitution

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

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

Expand

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

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

\int 1 d v = v

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

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

= - \frac{1}{6} (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= - \frac{1}{6} (sin (\frac{6}{x}) - \frac{2 sin ^{3} ( \frac{6}{x} )}{3} + \frac{sin ^{5} ( \frac{6}{x} )}{5})

Add a constant to the solution

= - \frac{1}{6} (sin (\frac{6}{x}) - \frac{2 sin ^{3} ( \frac{6}{x} )}{3} + \frac{sin ^{5} ( \frac{6}{x} )}{5}) + C

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What is the integral of (cos^5(6/x))/(x^2) ?
The integral of (cos^5(6/x))/(x^2) is -1/6 (sin(6/x)-(2sin^3(6/x))/3+(sin^5(6/x))/5)+C