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

The integral of (cos(x))/(sin^{19)(x)} is -1/18 csc^{18}(x)+C

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integral of (cos(x))/(sin^{19)(x)}

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

- \frac{1}{18} csc^{18} (x) + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int cos (x) sin^{- 19} (x) d x

Apply u-substitution

: \int \frac{1}{u ^{19}} d u

= \int \frac{1}{u ^{19}} d u

Apply the Power Rule

: - \frac{1}{18 u ^{18}}

= - \frac{1}{18 u ^{18}}

Substitute back

u = sin (x)

= - \frac{1}{18 sin ^{18} ( x )}

Simplify

- \frac{1}{18 sin ^{18} ( x )} : - \frac{1}{18} csc^{18} (x)

= - \frac{1}{18} csc^{18} (x)

Add a constant to the solution

= - \frac{1}{18} csc^{18} (x) + C

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What is the integral of (cos(x))/(sin^{19)(x)} ?
The integral of (cos(x))/(sin^{19)(x)} is -1/18 csc^{18}(x)+C