What is the integral of 1/(sin^2(3t)) ?

The integral of 1/(sin^2(3t)) is -1/3 cot(3t)+C

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integral of 1/(sin^2(3t))

\int \frac{1}{sin ^{2} ( 3 t )} d t

- \frac{1}{3} cot (3 t) + C

Solution steps

\int \frac{1}{sin ^{2} ( 3 t )} d t

Rewrite using trig identities

= \int csc^{2} (3 t) d t

Apply u-substitution

: \int csc^{2} (u) \frac{1}{3} d u

= \int csc^{2} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{3} (- cot (u))

Substitute back

u = 3 t

= \frac{1}{3} (- cot (3 t))

Simplify

= - \frac{1}{3} cot (3 t)

Add a constant to the solution

= - \frac{1}{3} cot (3 t) + C

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What is the integral of 1/(sin^2(3t)) ?
The integral of 1/(sin^2(3t)) is -1/3 cot(3t)+C