What is the integral of (9sec^2(x))/(tan(x)) ?

The integral of (9sec^2(x))/(tan(x)) is 9ln|tan(x)|+C

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integral of (9sec^2(x))/(tan(x))

\int \frac{9 sec ^{2} ( x )}{tan ( x )} d x

9 ln ∣ tan (x) ∣ + C

Solution steps

\int \frac{9 sec ^{2} ( x )}{tan ( x )} d x

Take the constant out:

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

= 9 \cdot \int \frac{sec ^{2} ( x )}{tan ( x )} d x

Apply u-substitution

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

= 9 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 ln ∣ u ∣

Substitute back

u = tan (x)

= 9 ln ∣ tan (x) ∣

Add a constant to the solution

= 9 ln ∣ tan (x) ∣ + C

x \to 2 lim (x + x)

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What is the integral of (9sec^2(x))/(tan(x)) ?
The integral of (9sec^2(x))/(tan(x)) is 9ln|tan(x)|+C