What is the integral of 2sin(θ)-sec^2(θ) ?

The integral of 2sin(θ)-sec^2(θ) is -2cos(θ)-tan(θ)+C

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integral of 2sin(θ)-sec^2(θ)

\int 2 sin (θ) - sec^{2} (θ) d θ

- 2 cos (θ) - tan (θ) + C

Solution steps

\int 2 sin (θ) - sec^{2} (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 2 sin (θ) d θ - \int sec^{2} (θ) d θ

\int 2 sin (θ) d θ = - 2 cos (θ)

\int sec^{2} (θ) d θ = tan (θ)

= - 2 cos (θ) - tan (θ)

Add a constant to the solution

= - 2 cos (θ) - tan (θ) + C

x \to + 0 lim (\frac{x}{arctan ( x )})

\int_{- 1}^{1} 3 e^{- 3 x} d x

5 e^{- 2 t}

y^{^{'}} = 12 t^{5} - 20 t^{4} + 2 - 6 t^{- 2}

y = t e^{- 3 t} sin (5 t)

What is the integral of 2sin(θ)-sec^2(θ) ?
The integral of 2sin(θ)-sec^2(θ) is -2cos(θ)-tan(θ)+C