What is the integral of 1/(sin(ax+b)) ?

The integral of 1/(sin(ax+b)) is 1/a ln|tan((ax+b)/2)|+C

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integral of 1/(sin(ax+b))

\int \frac{1}{sin ( a x + b )} d x

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

Solution steps

\int \frac{1}{sin ( a x + b )} d x

Apply u-substitution

: \int \frac{1}{a sin ( u )} d u

= \int \frac{1}{a sin ( u )} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{1}{sin ( u )} d u

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{a} ln ∣ v ∣

Substitute back

= \frac{1}{a} ln tan (\frac{a x + b}{2})

Add a constant to the solution

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

\frac{d y}{d x} = 6 x + y - 6

\int \frac{7 x ^{2} + 4 x + 7}{x ^{3} - 1} d x

\int_{- 3}^{3} (\frac{18}{9 + x ^{2}} - 1) d x

x \to - \infty lim (∣ x ∣)

2 \frac{d y}{d x} + 4 y = - 8 y^{3}

What is the integral of 1/(sin(ax+b)) ?
The integral of 1/(sin(ax+b)) is 1/a ln|tan((ax+b)/2)|+C