What is the integral of [6cos(x)-2sec(x)tan(x)] ?

The integral of [6cos(x)-2sec(x)tan(x)] is 6sin(x)-2sec(x)+C

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integral of [6cos(x)-2sec(x)tan(x)]

\int [6 cos (x) - 2 sec (x) tan (x)] d x

6 sin (x) - 2 sec (x) + C

Solution steps

\int 6 cos (x) - 2 sec (x) tan (x) d x

Apply the Sum Rule:

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

= \int 6 cos (x) d x - \int 2 sec (x) tan (x) d x

\int 6 cos (x) d x = 6 sin (x)

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

= 6 sin (x) - 2 sec (x)

Add a constant to the solution

= 6 sin (x) - 2 sec (x) + C

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

x \to 7 + lim (\frac{∣ x - 7 ∣}{x - 7})

lo g_{3} (x^{3} + 1)

\frac{d}{d x} (\frac{x}{3})

y = 2 x, [4, 9]

What is the integral of [6cos(x)-2sec(x)tan(x)] ?
The integral of [6cos(x)-2sec(x)tan(x)] is 6sin(x)-2sec(x)+C