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

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

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integral of (sec^2(θ)+sec(θ)tan(θ))

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

tan (θ) + sec (θ) + C

Solution steps

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

\int sec^{2} (θ) d θ = tan (θ)

\int sec (θ) tan (θ) d θ = sec (θ)

= tan (θ) + sec (θ)

Add a constant to the solution

= tan (θ) + sec (θ) + C

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

\frac{\partial}{\partial y} (2 - y - 2 x)

\frac{\partial}{\partial x} (x^{2} y sin (\frac{1}{x}))

x \to 0 lim (- x^{3})

f (x) = x^{2} + 6

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