What is the integral of (-cos(x))/(sin^9(x)) ?

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

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integral of (-cos(x))/(sin^9(x))

\int \frac{- cos ( x )}{sin ^{9} ( x )} d x

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

Solution steps

\int \frac{- cos ( x )}{sin ^{9} ( x )} d x

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{u ^{9}} d u

= - \int \frac{1}{u ^{9}} d u

Apply the Power Rule

: - \frac{1}{8 u ^{8}}

= - (- \frac{1}{8 u ^{8}})

Substitute back

u = sin (x)

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

Simplify

- (- \frac{1}{8 sin ^{8} ( x )}) : \frac{1}{8} csc^{8} (x)

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

Add a constant to the solution

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

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