What is the integral of (40-2/(sec(θ))) ?

The integral of (40-2/(sec(θ))) is 40θ-2sin(θ)+C

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integral of (40-2/(sec(θ)))

\int (40 - \frac{2}{sec ( θ )}) d θ

40 θ - 2 sin (θ) + C

Solution steps

\int 40 - \frac{2}{sec ( θ )} d θ

Use the basic trigonometric identity:

\frac{1}{sec ( x )} = cos (x)

= \int 40 - 2 cos (θ) d θ

Apply the Sum Rule:

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

= \int 40 d θ - \int 2 cos (θ) d θ

\int 40 d θ = 40 θ

\int 2 cos (θ) d θ = 2 sin (θ)

= 40 θ - 2 sin (θ)

Add a constant to the solution

= 40 θ - 2 sin (θ) + C

f (y) = e^{2 y}

\int_{- 2}^{1} - x^{2} - x + 2 d x

(5 + x^{8}) \frac{d y}{d x} = \frac{x ^{7}}{y}

\frac{d}{d x} (\frac{8 - 3 x}{x ^{2} - 6 x + 8})

x \to 0 lim (\frac{- 7 + x}{x ^{2}})

What is the integral of (40-2/(sec(θ))) ?
The integral of (40-2/(sec(θ))) is 40θ-2sin(θ)+C