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

The integral of 1/(sin(3x)) is 1/3 ln|tan((3x)/2)|+C

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integral of 1/(sin(3x))

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

\frac{1}{3} ln tan (\frac{3 x}{2}) + C

Solution steps

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

Rewrite using trig identities

= \int csc (3 x) d x

Apply u-substitution

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

= \int csc (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 (u) d u

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

Substitute back

= \frac{1}{3} ln tan (\frac{3 x}{2})

Add a constant to the solution

= \frac{1}{3} ln tan (\frac{3 x}{2}) + C

\int_{- 2}^{2} 3

(\frac{1}{x ^{2} - 1})^{^{'}}

f (x) = 4 - 3 x^{2}, (5, - 71)

\frac{- 0.5}{s ^{2} + s + 1}

\int x - 2 + \frac{3}{( 3 x - 5 ) ^{6}} d x

What is the integral of 1/(sin(3x)) ?
The integral of 1/(sin(3x)) is 1/3 ln|tan((3x)/2)|+C