What is the integral of (cos(pi/(x^{19)}))/(x^{20)} ?

The integral of (cos(pi/(x^{19)}))/(x^{20)} is -1/(19pi)sin(pi/(x^{19)})+C

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integral of (cos(pi/(x^{19)}))/(x^{20)}

\int \frac{cos ( \frac{π}{x ^{19}} )}{x ^{20}} d x

- \frac{1}{19 π} sin (\frac{π}{x ^{19}}) + C

Solution steps

\int \frac{cos ( \frac{π}{x ^{19}} )}{x ^{20}} d x

Apply u-substitution

: \int \frac{cos ( \frac{π}{u} )}{19 u ^{2}} d u

= \int \frac{cos ( \frac{π}{u} )}{19 u ^{2}} d u

Take the constant out:

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

= \frac{1}{19} \cdot \int \frac{cos ( \frac{π}{u} )}{u ^{2}} d u

Apply u-substitution

: \int - \frac{cos ( v )}{π} d v

= \frac{1}{19} \cdot \int - \frac{cos ( v )}{π} d v

Take the constant out:

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

= \frac{1}{19} (- \frac{1}{π} \cdot \int cos (v) d v)

Use the common integral:

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

= \frac{1}{19} (- \frac{1}{π} sin (v))

Substitute back

= \frac{1}{19} (- \frac{1}{π} sin (\frac{π}{x ^{19}}))

Simplify

\frac{1}{19} (- \frac{1}{π} sin (\frac{π}{x ^{19}})) : - \frac{1}{19 π} sin (\frac{π}{x ^{19}})

= - \frac{1}{19 π} sin (\frac{π}{x ^{19}})

Add a constant to the solution

= - \frac{1}{19 π} sin (\frac{π}{x ^{19}}) + C

\frac{d y}{d x} = \frac{1}{x} (y^{2} + 0.8 y)

\int (∣ x ∣) d x

\frac{d}{d x} (4 x^{2} + 3)

\int x 4 + x^{2} d x

x + \frac{4}{x}

What is the integral of (cos(pi/(x^{19)}))/(x^{20)} ?
The integral of (cos(pi/(x^{19)}))/(x^{20)} is -1/(19pi)sin(pi/(x^{19)})+C