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

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

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integral of 1/(9sin^2(x))

\int \frac{1}{9 sin ^{2} ( x )} d x

- \frac{1}{9} cot (x) + C

Solution steps

\int \frac{1}{9 sin ^{2} ( x )} d x

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} ( x )} d x

Use the common integral:

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

= \frac{1}{9} (- cot (x))

Simplify

= - \frac{1}{9} cot (x)

Add a constant to the solution

= - \frac{1}{9} cot (x) + C

x \to 0 lim (\frac{tan ( 4 x )}{tan ( 9 x )})

x \to 1 lim (\frac{( x ^{7 a} - 7 a x + 7 a - 1 )}{( x - 1 ) ^{2}})

\int \frac{1}{( x + 8 ) ^{2}} d x

\frac{d}{d x} (- y ln (x) - (1 - y) ln (1 - x))

x \to 1 + lim (x^{(\frac{1}{1 - x})})

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