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

The integral of cot(mx)csc^2(mx) is -1/(2m)csc^2(mx)+C

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integral of cot(mx)csc^2(mx)

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

- \frac{1}{2 m} csc^{2} (m x) + C

Solution steps

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

Apply u-substitution

: \int \frac{csc ^{2} ( u ) cot ( u )}{m} d u

= \int \frac{csc ^{2} ( u ) cot ( u )}{m} d u

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{2} d v

= \frac{1}{m} \cdot \int - \frac{1}{2} d v

Integral of a constant:

\int a d x = a x

= \frac{1}{m} (- \frac{1}{2}) v

Substitute back

= \frac{1}{m} (- \frac{1}{2}) csc^{2} (m x)

Simplify

\frac{1}{m} (- \frac{1}{2}) csc^{2} (m x) : - \frac{1}{2 m} csc^{2} (m x)

= - \frac{1}{2 m} csc^{2} (m x)

Add a constant to the solution

= - \frac{1}{2 m} csc^{2} (m x) + C

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What is the integral of cot(mx)csc^2(mx) ?
The integral of cot(mx)csc^2(mx) is -1/(2m)csc^2(mx)+C