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

The integral of 2tan(x)sec(x) is 2sec(x)+C

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integral of 2tan(x)sec(x)

\int 2 tan (x) sec (x) d x

2 sec (x) + C

Solution steps

\int 2 tan (x) sec (x) d x

Take the constant out:

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

= 2 \cdot \int tan (x) sec (x) d x

Apply u-substitution

: \int 1 d u

= 2 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot u

Substitute back

u = sec (x)

= 2 \cdot 1 \cdot sec (x)

Simplify

= 2 sec (x)

Add a constant to the solution

= 2 sec (x) + C

(16, 13), (- 8, - 17)

\int_{4}^{\infty} e^{- 9 x} d x

\int 16 - 9 x^{2} d x

\frac{\partial}{\partial x} (3 x + y^{2})

\frac{d}{d x} (\frac{x}{a ^{2} - x ^{2}})

What is the integral of 2tan(x)sec(x) ?
The integral of 2tan(x)sec(x) is 2sec(x)+C