What is the integral of-1/(sin^2(rx)) ?

The integral of-1/(sin^2(rx)) is 1/r cot(rx)+C

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integral of-1/(sin^2(rx))

\int - \frac{1}{sin ^{2} ( r x )} d x

\frac{1}{r} cot (r x) + C

Solution steps

\int - \frac{1}{sin ^{2} ( r x )} d x

Take the constant out:

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

= - \int \frac{1}{sin ^{2} ( r x )} d x

Rewrite using trig identities

= - \int csc^{2} (r x) d x

Apply u-substitution

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

= - \int \frac{csc ^{2} ( u )}{r} d u

Take the constant out:

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

= - \frac{1}{r} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= - \frac{1}{r} (- cot (u))

Substitute back

u = r x

= - \frac{1}{r} (- cot (r x))

Simplify

= \frac{1}{r} cot (r x)

Add a constant to the solution

= \frac{1}{r} cot (r x) + C

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What is the integral of-1/(sin^2(rx)) ?
The integral of-1/(sin^2(rx)) is 1/r cot(rx)+C