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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{u}{5} d u

= 10 \cdot \int - \frac{u}{5} d u

Take the constant out:

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

= 10 (- \frac{1}{5} \cdot \int u d u)

Apply the Power Rule

: \frac{u ^{2}}{2}

= 10 (- \frac{1}{5} \cdot \frac{u ^{2}}{2})

Substitute back

u = cot (5 x)

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

Simplify

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

= - cot^{2} (5 x)

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{sin ( x + y )}{y ^{2}})

4 \frac{d y}{d x} - 12 x^{2} = 0

y = \frac{80}{x ^{6}}

y = 3 x + 2 cos (x)

\int 6 e^{4 x} d x

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