What is the integral of (csc(5/(x^3)))/(x^4) ?

The integral of (csc(5/(x^3)))/(x^4) is -1/15 ln|tan(5/(2x^3))|+C

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integral of (csc(5/(x^3)))/(x^4)

\int \frac{csc ( \frac{5}{x ^{3}} )}{x ^{4}} d x

- \frac{1}{15} ln tan (\frac{5}{2 x ^{3}}) + C

Solution steps

\int \frac{csc ( \frac{5}{x ^{3}} )}{x ^{4}} d x

Apply u-substitution

: \int - \frac{csc ( u )}{15} d u

= \int - \frac{csc ( u )}{15} d u

Take the constant out:

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

= - \frac{1}{15} \cdot \int csc (u) d u

Apply u-substitution

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

= - \frac{1}{15} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{15} ln ∣ v ∣

Substitute back

= - \frac{1}{15} ln tan (\frac{\frac{5}{x ^{3}}}{2})

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= - \frac{1}{15} ln tan (\frac{5}{2 x ^{3}})

Add a constant to the solution

= - \frac{1}{15} ln tan (\frac{5}{2 x ^{3}}) + C

f (x) = x^{2} + x, (2, - 3)

x \to 8 + lim (\frac{1}{( x - 8 )})

(7 x + 3 y) d x + (3 x - 8 y 3) d y = 0

\frac{k x + a}{b x + c}

\frac{\partial}{\partial z} (z sin (y))

What is the integral of (csc(5/(x^3)))/(x^4) ?
The integral of (csc(5/(x^3)))/(x^4) is -1/15 ln|tan(5/(2x^3))|+C