What is the integral of cot^5(x)csc^2(x) ?

The integral of cot^5(x)csc^2(x) is -(cot^6(x))/6+C

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integral of cot^5(x)csc^2(x)

\int cot^{5} (x) csc^{2} (x) d x

- \frac{cot ^{6} ( x )}{6} + C

Solution steps

\int cot^{5} (x) csc^{2} (x) d x

Rewrite using trig identities

= \int \frac{cos ^{5} ( x )}{sin ^{7} ( x )} d x

Apply Trigonometric Substitution

: \int - u^{5} d u

= \int - u^{5} d u

Take the constant out:

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

= - \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= - \frac{u ^{6}}{6}

Substitute back

u = cot (x)

= - \frac{cot ^{6} ( x )}{6}

Add a constant to the solution

= - \frac{cot ^{6} ( x )}{6} + C

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x \to 0 lim ((1 + \frac{4}{x ^{4}})^{x})

(\frac{- 9 x ^{2} + 12 x + 24}{( x ^{2} - 5 x - 6 ) ^{2}})^{^{'}}

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What is the integral of cot^5(x)csc^2(x) ?
The integral of cot^5(x)csc^2(x) is -(cot^6(x))/6+C