What is the integral of (3sec^2(t))/(6+3tan(t)) ?

The integral of (3sec^2(t))/(6+3tan(t)) is ln|6+3tan(t)|+C

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integral of (3sec^2(t))/(6+3tan(t))

\int \frac{3 sec ^{2} ( t )}{6 + 3 tan ( t )} d t

ln ∣ 6 + 3 tan (t) ∣ + C

Solution steps

\int \frac{3 sec ^{2} ( t )}{6 + 3 tan ( t )} d t

Take the constant out:

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

= 3 \cdot \int \frac{sec ^{2} ( t )}{6 + 3 tan ( t )} d t

Apply u-substitution

= 3 \cdot \int \frac{1}{3 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 6 + 3 tan (t)

= 3 \cdot \frac{1}{3} ln ∣ 6 + 3 tan (t) ∣

Simplify

3 \cdot \frac{1}{3} ln ∣ 6 + 3 tan (t) ∣ : ln ∣ 6 + 3 tan (t) ∣

= ln ∣ 6 + 3 tan (t) ∣

Add a constant to the solution

= ln ∣ 6 + 3 tan (t) ∣ + C

What is the integral of (3sec^2(t))/(6+3tan(t)) ?
The integral of (3sec^2(t))/(6+3tan(t)) is ln|6+3tan(t)|+C