What is the integral of (tan(4x)+sec(4x))^2 ?

The integral of (tan(4x)+sec(4x))^2 is 1/2 (-2x+tan(4x)+sec(4x))+C

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integral of (tan(4x)+sec(4x))^2

\int (tan (4 x) + sec (4 x))^{2} d x

\frac{1}{2} (- 2 x + tan (4 x) + sec (4 x)) + C

Solution steps

\int (tan (4 x) + sec (4 x))^{2} d x

Apply u-substitution

: \int (tan (u) + sec (u))^{2} \frac{1}{4} d u

= \int (tan (u) + sec (u))^{2} \frac{1}{4} 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 (tan (u) + sec (u))^{2} d u

Expand

(tan (u) + sec (u))^{2} : tan^{2} (u) + 2 tan (u) sec (u) + sec^{2} (u)

= \frac{1}{4} \cdot \int tan^{2} (u) + 2 tan (u) sec (u) + sec^{2} (u) d u

Apply the Sum Rule:

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

= \frac{1}{4} (\int tan^{2} (u) d u + \int 2 tan (u) sec (u) d u + \int sec^{2} (u) d u)

\int tan^{2} (u) d u = - u + tan (u)

\int 2 tan (u) sec (u) d u = 2 sec (u)

\int sec^{2} (u) d u = tan (u)

= \frac{1}{4} (- u + tan (u) + 2 sec (u) + tan (u))

Substitute back

u = 4 x

= \frac{1}{4} (- 4 x + tan (4 x) + 2 sec (4 x) + tan (4 x))

Simplify

\frac{1}{4} (- 4 x + tan (4 x) + 2 sec (4 x) + tan (4 x)) : \frac{1}{2} (- 2 x + tan (4 x) + sec (4 x))

= \frac{1}{2} (- 2 x + tan (4 x) + sec (4 x))

Add a constant to the solution

= \frac{1}{2} (- 2 x + tan (4 x) + sec (4 x)) + C

\frac{\partial}{\partial x} (e^{3 x - y})

f (x) = 3 e^{1 - x^{2}}, (1, 3)

n = 0 \sum \infty - \frac{1}{5 ^{n}}

11 x^{2} y + \frac{d y}{d x} = - 6 x^{2}

x \to - 5 lim (\frac{x + 5}{x + 5})

What is the integral of (tan(4x)+sec(4x))^2 ?
The integral of (tan(4x)+sec(4x))^2 is 1/2 (-2x+tan(4x)+sec(4x))+C