What is the integral of 2csc(x)sec(x) ?

The integral of 2csc(x)sec(x) is ln(tan^2(x))+C

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integral of 2csc(x)sec(x)

\int 2 csc (x) sec (x) d x

ln (tan^{2} (x)) + C

Solution steps

\int 2 csc (x) sec (x) d x

Take the constant out:

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

= 2 \cdot \int csc (x) sec (x) d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 1} d u

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

Apply u-substitution

: \int \frac{1}{2 v} d v

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

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

= 2 \cdot \frac{1}{2} ln sec^{2} (x) - 1

Simplify

2 \cdot \frac{1}{2} ln sec^{2} (x) - 1 : ln (tan^{2} (x))

= ln (tan^{2} (x))

Add a constant to the solution

= ln (tan^{2} (x)) + C

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What is the integral of 2csc(x)sec(x) ?
The integral of 2csc(x)sec(x) is ln(tan^2(x))+C