What is the integral of 1/(9sin^2(θ)) ?

The integral of 1/(9sin^2(θ)) is -1/9 cot(θ)+C

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integral of 1/(9sin^2(θ))

\int \frac{1}{9 sin ^{2} ( θ )} d θ

- \frac{1}{9} cot (θ) + C

Solution steps

\int \frac{1}{9 sin ^{2} ( θ )} d θ

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{sin ^{2} ( θ )} d θ

Use the common integral:

\int \frac{1}{sin ^{2} ( θ )} d θ = - cot (θ)

= \frac{1}{9} (- cot (θ))

Simplify

= - \frac{1}{9} cot (θ)

Add a constant to the solution

= - \frac{1}{9} cot (θ) + C

\int_{0}^{π} sin^{5} (\frac{x}{2}) d x

t \to π lim (sin (\frac{t}{6}))

\frac{d y}{d x} = 2 y 16 - x^{2}, y (0) = 1

\frac{d}{d x} (ln (16 - x^{2}))

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

What is the integral of 1/(9sin^2(θ)) ?
The integral of 1/(9sin^2(θ)) is -1/9 cot(θ)+C