What is the integral of (tan(5x)+sec(5x))^2 ?

The integral of (tan(5x)+sec(5x))^2 is 1/5 (-5x+2tan(5x)+2sec(5x))+C

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integral of (tan(5x)+sec(5x))^2

\int (tan (5 x) + sec (5 x))^{2} d x

\frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x)) + C

Solution steps

\int (tan (5 x) + sec (5 x))^{2} d x

Apply u-substitution

: \int (tan (u) + sec (u))^{2} \frac{1}{5} d u

= \int (tan (u) + sec (u))^{2} \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \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}{5} \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}{5} (\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}{5} (- u + tan (u) + 2 sec (u) + tan (u))

Substitute back

u = 5 x

= \frac{1}{5} (- 5 x + tan (5 x) + 2 sec (5 x) + tan (5 x))

Simplify

\frac{1}{5} (- 5 x + tan (5 x) + 2 sec (5 x) + tan (5 x)) : \frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x))

= \frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x))

Add a constant to the solution

= \frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x)) + C

\frac{d y}{d x} = (x^{2} - 3 x)^{3}

\frac{1}{x ^{3}}

\int_{0}^{2} - 5 x^{3} x^{2} + 4 d x

\int 5 θ cos (θ) d θ

a x + 1

What is the integral of (tan(5x)+sec(5x))^2 ?
The integral of (tan(5x)+sec(5x))^2 is 1/5 (-5x+2tan(5x)+2sec(5x))+C