What is the integral of (sin(1/(x^8)))/(x^9) ?

The integral of (sin(1/(x^8)))/(x^9) is 1/8 cos(1/(x^8))+C

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integral of (sin(1/(x^8)))/(x^9)

\int \frac{sin ( \frac{1}{x ^{8}} )}{x ^{9}} d x

\frac{1}{8} cos (\frac{1}{x ^{8}}) + C

Solution steps

\int \frac{sin ( \frac{1}{x ^{8}} )}{x ^{9}} d x

Apply u-substitution

: \int \frac{sin ( \frac{1}{u} )}{8 u ^{2}} d u

= \int \frac{sin ( \frac{1}{u} )}{8 u ^{2}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{sin ( \frac{1}{u} )}{u ^{2}} d u

Apply u-substitution

: \int - sin (v) d v

= \frac{1}{8} \cdot \int - sin (v) d v

Take the constant out:

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

= \frac{1}{8} (- \int sin (v) d v)

Use the common integral:

\int sin (v) d v = - cos (v)

= \frac{1}{8} (- (- cos (v)))

Substitute back

= \frac{1}{8} (- (- cos (\frac{1}{x ^{8}})))

Simplify

= \frac{1}{8} cos (\frac{1}{x ^{8}})

Add a constant to the solution

= \frac{1}{8} cos (\frac{1}{x ^{8}}) + C

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What is the integral of (sin(1/(x^8)))/(x^9) ?
The integral of (sin(1/(x^8)))/(x^9) is 1/8 cos(1/(x^8))+C