What is the integral of (13(tan(x))^2)/(sec(x)) ?

The integral of (13(tan(x))^2)/(sec(x)) is 13(ln|tan(x)+sec(x)|-sin(x))+C

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integral of (13(tan(x))^2)/(sec(x))

\int \frac{13 ( tan ( x ) ) ^{2}}{sec ( x )} d x

13 (ln ∣ tan (x) + sec (x) ∣ - sin (x)) + C

Solution steps

\int \frac{13 ( tan ( x ) ) ^{2}}{sec ( x )} d x

Take the constant out:

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

= 13 \cdot \int \frac{( tan ( x ) ) ^{2}}{sec ( x )} d x

Simplify

\frac{( tan ( x ) ) ^{2}}{sec ( x )} : sin (x) tan (x)

= 13 \cdot \int sin (x) tan (x) d x

Rewrite using trig identities

= 13 \cdot \int \frac{1 - cos ^{2} ( x )}{cos ( x )} d x

Expand

\frac{1 - cos ^{2} ( x )}{cos ( x )} : \frac{1}{cos ( x )} - cos (x)

= 13 \cdot \int \frac{1}{cos ( x )} - cos (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 13 (\int \frac{1}{cos ( x )} d x - \int cos (x) d x)

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

\int cos (x) d x = sin (x)

= 13 (ln ∣ tan (x) + sec (x) ∣ - sin (x))

Add a constant to the solution

= 13 (ln ∣ tan (x) + sec (x) ∣ - sin (x)) + C

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What is the integral of (13(tan(x))^2)/(sec(x)) ?
The integral of (13(tan(x))^2)/(sec(x)) is 13(ln|tan(x)+sec(x)|-sin(x))+C