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

The integral of (θ^6+sec^2(θ)) is (θ^7)/7+tan(θ)+C

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integral of (θ^6+sec^2(θ))

\int (θ^{6} + sec^{2} (θ)) d θ

\frac{θ ^{7}}{7} + tan (θ) + C

Solution steps

\int θ^{6} + 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^{6} d θ + \int sec^{2} (θ) d θ

\int θ^{6} d θ = \frac{θ ^{7}}{7}

\int sec^{2} (θ) d θ = tan (θ)

= \frac{θ ^{7}}{7} + tan (θ)

Add a constant to the solution

= \frac{θ ^{7}}{7} + tan (θ) + C

28 - x^{2}

\int \frac{y}{3 + x y} d x

\int \frac{1}{( 3 x + 2 ) ^{3}} d x

(- 4 x^{2} + 16)^{2}

n = 1 \sum \infty (- \frac{3}{4})^{n - 1}

What is the integral of (θ^6+sec^2(θ)) ?
The integral of (θ^6+sec^2(θ)) is (θ^7)/7+tan(θ)+C