What is the integral of (e^0)/((sin(3x))^2) ?

The integral of (e^0)/((sin(3x))^2) is -1/3 cot(3x)+C

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integral of (e^0)/((sin(3x))^2)

\int \frac{e ^{0}}{( sin ( 3 x ) ) ^{2}} d x

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

Solution steps

\int \frac{e ^{0}}{( sin ( 3 x ) ) ^{2}} d x

Take the constant out:

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

= e^{0} \cdot \int \frac{1}{( sin ( 3 x ) ) ^{2}} d x

Simplify

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

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

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

Apply u-substitution

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

= 1 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = 3 x

= 1 \cdot \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

n = 1 \sum \infty \frac{1}{2 ^{n} n}

\int \frac{x}{1 - x ^{3}} d x

\frac{d}{d x} (a (2 e^{- 5 x} x - 5 e^{- 5 x} x^{2}))

\int \frac{1}{4 - x} d x

\frac{d}{d x} (\frac{x + 1}{x})

What is the integral of (e^0)/((sin(3x))^2) ?
The integral of (e^0)/((sin(3x))^2) is -1/3 cot(3x)+C