What is the integral of (8sec^2(4t))/(3+2tan(4t)) ?

The integral of (8sec^2(4t))/(3+2tan(4t)) is ln|3+2tan(4t)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8sec^2(4t))/(3+2tan(4t))

\int \frac{8 sec ^{2} ( 4 t )}{3 + 2 tan ( 4 t )} d t

ln ∣ 3 + 2 tan (4 t) ∣ + C

Solution steps

\int \frac{8 sec ^{2} ( 4 t )}{3 + 2 tan ( 4 t )} d t

Take the constant out:

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

= 8 \cdot \int \frac{sec ^{2} ( 4 t )}{3 + 2 tan ( 4 t )} d t

Apply u-substitution

: \int \frac{1}{8 u} d u

= 8 \cdot \int \frac{1}{8 u} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 3 + 2 tan (4 t)

= 8 \cdot \frac{1}{8} ln ∣ 3 + 2 tan (4 t) ∣

Simplify

8 \cdot \frac{1}{8} ln ∣ 3 + 2 tan (4 t) ∣ : ln ∣ 3 + 2 tan (4 t) ∣

= ln ∣ 3 + 2 tan (4 t) ∣

Add a constant to the solution

= ln ∣ 3 + 2 tan (4 t) ∣ + C

\frac{x}{x ^{2} + 16}

\int \frac{x ^{2} - 4}{x - 2} d x

f (x) = (3 x^{2} + 2)^{2} (x^{2} - 5 x)^{3}

\int x^{2} ln (x^{3}) d x

x \to - 1 - lim (x^{3} + 2 x^{2} + 1)

What is the integral of (8sec^2(4t))/(3+2tan(4t)) ?
The integral of (8sec^2(4t))/(3+2tan(4t)) is ln|3+2tan(4t)|+C