What is the integral of (sec^2(θ)+6) ?

The integral of (sec^2(θ)+6) is tan(θ)+6θ+C

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integral of (sec^2(θ)+6)

\int (sec^{2} (θ) + 6) d θ

tan (θ) + 6 θ + C

Solution steps

\int sec^{2} (θ) + 6 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 6 d θ

\int sec^{2} (θ) d θ = tan (θ)

\int 6 d θ = 6 θ

= tan (θ) + 6 θ

Add a constant to the solution

= tan (θ) + 6 θ + C

1 - 2 sin^{2} (x)

y = (\frac{x + 3}{x - 1})^{2}, (5, 4)

f (x) = sin (x) + arcsin (x)

\frac{d}{d x} (lo g_{2} (5 x))

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

What is the integral of (sec^2(θ)+6) ?
The integral of (sec^2(θ)+6) is tan(θ)+6θ+C