What is the integral of xcsc(4x^2) ?

The integral of xcsc(4x^2) is 1/8 ln|tan(2x^2)|+C

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integral of xcsc(4x^2)

\int x csc (4 x^{2}) d x

\frac{1}{8} ln tan (2 x^{2}) + C

Solution steps

\int x csc (4 x^{2}) d x

Apply u-substitution

: \int \frac{csc ( u )}{8} d u

= \int \frac{csc ( u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int csc (u) d u

Apply u-substitution

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

= \frac{1}{8} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{8} ln ∣ v ∣

Substitute back

= \frac{1}{8} ln tan (\frac{4 x ^{2}}{2})

Simplify

\frac{1}{8} ln tan (\frac{4 x ^{2}}{2}) : \frac{1}{8} ln tan (2 x^{2})

= \frac{1}{8} ln tan (2 x^{2})

Add a constant to the solution

= \frac{1}{8} ln tan (2 x^{2}) + C

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What is the integral of xcsc(4x^2) ?
The integral of xcsc(4x^2) is 1/8 ln|tan(2x^2)|+C