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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 x

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{sec ( y )}{( x - 3 ) ^{2}}

\frac{d y}{d x} - 2 x y = x y^{2}

\frac{7 - sec ( x )}{tan ( x )}

\int_{0}^{1} 8 cos (x^{2}) d x

f (x) = 2 + 2 e^{- x}, at x = ln (5)

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