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

The integral of (cos(3x))/(x^3) is 9(-(cos(3x))/(18x^2)-1/2 (-(sin(3x))/(3x)+\Ci(3x)))+C

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integral of (cos(3x))/(x^3)

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

9 (- \frac{cos ( 3 x )}{18 x ^{2}} - \frac{1}{2} (- \frac{sin ( 3 x )}{3 x} + Ci (3 x))) + C

Solution steps

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

Apply u-substitution

: \int \frac{9 cos ( u )}{u ^{3}} d u

= \int \frac{9 cos ( u )}{u ^{3}} d u

Take the constant out:

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

= 9 \cdot \int \frac{cos ( u )}{u ^{3}} d u

Apply Integration By Parts

: - \frac{cos ( u )}{2 u ^{2}} - \int \frac{sin ( u )}{2 u ^{2}} d u

= 9 (- \frac{cos ( u )}{2 u ^{2}} - \int \frac{sin ( u )}{2 u ^{2}} d u)

\int \frac{sin ( u )}{2 u ^{2}} d u = \frac{1}{2} (- \frac{sin ( u )}{u} + Ci (u))

= 9 (- \frac{cos ( u )}{2 u ^{2}} - \frac{1}{2} (- \frac{sin ( u )}{u} + Ci (u)))

Substitute back

u = 3 x

= 9 (- \frac{cos ( 3 x )}{2 ( 3 x ) ^{2}} - \frac{1}{2} (- \frac{sin ( 3 x )}{3 x} + Ci (3 x)))

Simplify

9 (- \frac{cos ( 3 x )}{2 ( 3 x ) ^{2}} - \frac{1}{2} (- \frac{sin ( 3 x )}{3 x} + Ci (3 x))) : 9 (- \frac{cos ( 3 x )}{18 x ^{2}} - \frac{1}{2} (- \frac{sin ( 3 x )}{3 x} + Ci (3 x)))

= 9 (- \frac{cos ( 3 x )}{18 x ^{2}} - \frac{1}{2} (- \frac{sin ( 3 x )}{3 x} + Ci (3 x)))

Add a constant to the solution

= 9 (- \frac{cos ( 3 x )}{18 x ^{2}} - \frac{1}{2} (- \frac{sin ( 3 x )}{3 x} + Ci (3 x))) + C

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What is the integral of (cos(3x))/(x^3) ?
The integral of (cos(3x))/(x^3) is 9(-(cos(3x))/(18x^2)-1/2 (-(sin(3x))/(3x)+\Ci(3x)))+C