What is the integral of-1/4 csc(8t) ?

The integral of-1/4 csc(8t) is -1/32 ln|tan(4t)|+C

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integral of-1/4 csc(8t)

\int - \frac{1}{4} csc (8 t) d t

- \frac{1}{32} ln ∣ tan (4 t) ∣ + C

Solution steps

\int - \frac{1}{4} csc (8 t) d t

Take the constant out:

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

= - \frac{1}{4} \cdot \int csc (8 t) d t

Apply u-substitution

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

= - \frac{1}{4} \cdot \int csc (u) \frac{1}{8} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \frac{1}{8} \cdot \int csc (u) d u

Apply u-substitution

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

= - \frac{1}{4} \cdot \frac{1}{8} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{4} \cdot \frac{1}{8} ln ∣ v ∣

Substitute back

= - \frac{1}{4} \cdot \frac{1}{8} ln tan (\frac{8 t}{2})

Simplify

- \frac{1}{4} \cdot \frac{1}{8} ln tan (\frac{8 t}{2}) : - \frac{1}{32} ln ∣ tan (4 t) ∣

= - \frac{1}{32} ln ∣ tan (4 t) ∣

Add a constant to the solution

= - \frac{1}{32} ln ∣ tan (4 t) ∣ + C

\frac{\partial}{\partial x} (x^{2} + y e^{x})

\int \frac{11 - 2 x}{x ^{2} + 9 x + 14} d x

\frac{2 x}{3 - cot ( x )}

x \to 3 lim ((x^{2} - 9)^{x - 3})

x \to \infty lim (\frac{x}{9})

What is the integral of-1/4 csc(8t) ?
The integral of-1/4 csc(8t) is -1/32 ln|tan(4t)|+C