What is the integral of (θ^4+sec^2(θ)) ?

The integral of (θ^4+sec^2(θ)) is (θ^5)/5+tan(θ)+C

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integral of (θ^4+sec^2(θ))

\int (θ^{4} + sec^{2} (θ)) d θ

\frac{θ ^{5}}{5} + tan (θ) + C

Solution steps

\int θ^{4} + 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^{4} d θ + \int sec^{2} (θ) d θ

\int θ^{4} d θ = \frac{θ ^{5}}{5}

\int sec^{2} (θ) d θ = tan (θ)

= \frac{θ ^{5}}{5} + tan (θ)

Add a constant to the solution

= \frac{θ ^{5}}{5} + tan (θ) + C

y = sin (2) (x), y = sin (3) (x), 0 \leq x \leq π

x - \frac{1}{x}

t e^{- t} sin (2 t)

y^{^{'}} - \frac{4}{t} y = t^{5} e^{t}

x \to 1 - lim (\frac{x - 3}{x - 1})

What is the integral of (θ^4+sec^2(θ)) ?
The integral of (θ^4+sec^2(θ)) is (θ^5)/5+tan(θ)+C