What is the integral of 10csc^2(-5x)cot^4(-5x) ?

The integral of 10csc^2(-5x)cot^4(-5x) is -2/5 cot^5(5x)+C

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integral of 10csc^2(-5x)cot^4(-5x)

\int 10 csc^{2} (- 5 x) cot^{4} (- 5 x) d x

- \frac{2}{5} cot^{5} (5 x) + C

Solution steps

\int 10 csc^{2} (- 5 x) cot^{4} (- 5 x) d x

Take the constant out:

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

= 10 \cdot \int csc^{2} (- 5 x) cot^{4} (- 5 x) d x

Simplify

csc^{2} (- 5 x) cot^{4} (- 5 x) : cot^{4} (5 x) csc^{2} (5 x)

= 10 \cdot \int cot^{4} (5 x) csc^{2} (5 x) d x

Apply u-substitution

= 10 \cdot \int cot^{4} (u) csc^{2} (u) \frac{1}{5} d u

Take the constant out:

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

= 10 \cdot \frac{1}{5} \cdot \int cot^{4} (u) csc^{2} (u) d u

Apply u-substitution

= 10 \cdot \frac{1}{5} \cdot \int - v^{4} d v

Take the constant out:

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

= 10 \cdot \frac{1}{5} (- \int v^{4} d v)

Apply the Power Rule

= 10 \cdot \frac{1}{5} (- \frac{v ^{5}}{5})

Substitute back

= 10 \cdot \frac{1}{5} (- \frac{cot ^{5} ( 5 x )}{5})

Simplify

10 \cdot \frac{1}{5} (- \frac{cot ^{5} ( 5 x )}{5}) : - \frac{2}{5} cot^{5} (5 x)

= - \frac{2}{5} cot^{5} (5 x)

Add a constant to the solution

= - \frac{2}{5} cot^{5} (5 x) + C

What is the integral of 10csc^2(-5x)cot^4(-5x) ?
The integral of 10csc^2(-5x)cot^4(-5x) is -2/5 cot^5(5x)+C