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

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

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integral of (2+tan^2(θ))

\int (2 + tan^{2} (θ)) d θ

tan (θ) + θ + C

Solution steps

\int 2 + tan^{2} (θ) d θ

Rewrite using trig identities

= \int sec^{2} (θ) + 1 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 1 d θ

\int sec^{2} (θ) d θ = tan (θ)

\int 1 d θ = θ

= tan (θ) + θ

Add a constant to the solution

= tan (θ) + θ + C

\frac{d}{d x} (\frac{5 x}{x ^{2} + 1})

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

f (x) = \frac{8 x ^{3} - 3 x + 2}{3 x - x ^{4}}

(cos (x^{2}) \cdot 2 x)^{^{'}}

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + x y))

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