What is the integral of (sec^2(θ)+5) ?

The integral of (sec^2(θ)+5) is tan(θ)+5θ+C

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integral of (sec^2(θ)+5)

\int (sec^{2} (θ) + 5) d θ

tan (θ) + 5 θ + C

Solution steps

\int sec^{2} (θ) + 5 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 5 d θ

\int sec^{2} (θ) d θ = tan (θ)

\int 5 d θ = 5 θ

= tan (θ) + 5 θ

Add a constant to the solution

= tan (θ) + 5 θ + C

\frac{d y}{d x} + y^{9} x + 7 y = 0

\int \frac{2 x ^{2} + 3}{x ^{3}} d x

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

n = 1 \sum \infty \frac{2}{3 ^{n + 1}}

\frac{1}{S}

What is the integral of (sec^2(θ)+5) ?
The integral of (sec^2(θ)+5) is tan(θ)+5θ+C