What is the integral of (csc(θ)cot(θ))/(5-4csc(θ)) ?

The integral of (csc(θ)cot(θ))/(5-4csc(θ)) is 1/4 ln|5-4csc(θ)|+C

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integral of (csc(θ)cot(θ))/(5-4csc(θ))

\int \frac{csc ( θ ) cot ( θ )}{5 - 4 csc ( θ )} d θ

\frac{1}{4} ln ∣ 5 - 4 csc (θ) ∣ + C

Solution steps

\int \frac{csc ( θ ) cot ( θ )}{5 - 4 csc ( θ )} d θ

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 5 - 4 csc (θ)

= \frac{1}{4} ln ∣ 5 - 4 csc (θ) ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 5 - 4 csc (θ) ∣ + C

f (x) = e^{6 x}

(x + 2) d t = 2 t + 3 d x

\frac{\partial}{\partial x} (\frac{x ^{2}}{4})

\frac{\partial}{\partial x} (x^{6} y^{4} - x^{5} y^{3})

\int (\frac{1}{x ^{2} - 4}) d x

What is the integral of (csc(θ)cot(θ))/(5-4csc(θ)) ?
The integral of (csc(θ)cot(θ))/(5-4csc(θ)) is 1/4 ln|5-4csc(θ)|+C