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

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

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integral of 2/(sin(2x))

\int \frac{2}{sin ( 2 x )} d x

ln ∣ tan (x) ∣ + C

Solution steps

\int \frac{2}{sin ( 2 x )} d x

Take the constant out:

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

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

Rewrite using trig identities

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply Trigonometric Substitution

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = tan (x)

= 2 \cdot \frac{1}{2} ln ∣ tan (x) ∣

Simplify

2 \cdot \frac{1}{2} ln ∣ tan (x) ∣ : ln ∣ tan (x) ∣

= ln ∣ tan (x) ∣

Add a constant to the solution

= ln ∣ tan (x) ∣ + C

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

\frac{d ^{2}}{d x ^{2}} (\frac{e ^{x}}{cos ( x )})

\frac{\partial}{\partial y} (x z + x yz)

(- 2, - 5), (- 3, 5)

\int (x - 25 - x^{2}) d x

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