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

The integral of (θ^2+sec^2(θ)) is (θ^3)/3+tan(θ)+C

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integral of (θ^2+sec^2(θ))

\int (θ^{2} + sec^{2} (θ)) d θ

\frac{θ ^{3}}{3} + tan (θ) + C

Solution steps

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

\int θ^{2} d θ = \frac{θ ^{3}}{3}

\int sec^{2} (θ) d θ = tan (θ)

= \frac{θ ^{3}}{3} + tan (θ)

Add a constant to the solution

= \frac{θ ^{3}}{3} + tan (θ) + C

\int cos^{4} (\frac{x}{2}) d x

2 t^{2}

\int_{0}^{9} (9 - y + 2)^{2} d y

9 x y^{^{'}} = 2 x e^{x} - 9 y + 6 x^{2}

\frac{2 s + 6}{s ^{2} + 4}

What is the integral of (θ^2+sec^2(θ)) ?
The integral of (θ^2+sec^2(θ)) is (θ^3)/3+tan(θ)+C