What is the integral of (θ^8+sec^2(θ)) ?

The integral of (θ^8+sec^2(θ)) is (θ^9)/9+tan(θ)+C

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integral of (θ^8+sec^2(θ))

\int (θ^{8} + sec^{2} (θ)) d θ

\frac{θ ^{9}}{9} + tan (θ) + C

Solution steps

\int θ^{8} + 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^{8} d θ + \int sec^{2} (θ) d θ

\int θ^{8} d θ = \frac{θ ^{9}}{9}

\int sec^{2} (θ) d θ = tan (θ)

= \frac{θ ^{9}}{9} + tan (θ)

Add a constant to the solution

= \frac{θ ^{9}}{9} + tan (θ) + C

sin (x), \frac{2 π}{3}

\int (4 x - sin (x)) d x

\int_{5}^{1} (- 4 x + 1) d x

x \to 3 - lim (2)

f^{^{'}} = f^{3}

What is the integral of (θ^8+sec^2(θ)) ?
The integral of (θ^8+sec^2(θ)) is (θ^9)/9+tan(θ)+C