What is the integral from sec(θ) to 2 of 3r^2sin(θ) ?

The integral from sec(θ) to 2 of 3r^2sin(θ) is sin(θ)(8-sec^3(θ))

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integral from sec(θ) to 2 of 3r^2sin(θ)

\int_{s e c (θ)}^{2} 3 r^{2} sin (θ) d r

sin (θ) (8 - sec^{3} (θ))

Solution steps

\int_{s e c (θ)}^{2} 3 r^{2} sin (θ) d r

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 sin (θ) \cdot \int_{s e c (θ)}^{2} r^{2} d r

Apply the Power Rule

: [\frac{r ^{3}}{3}]_{s e c (θ)}^{2}

= 3 sin (θ) [\frac{r ^{3}}{3}]_{s e c (θ)}^{2}

Compute the boundaries:

\frac{8}{3} - \frac{sec ^{3} ( θ )}{3}

= 3 sin (θ) (\frac{8}{3} - \frac{sec ^{3} ( θ )}{3})

Simplify

= sin (θ) (8 - sec^{3} (θ))

\int_{1}^{2} \frac{1}{x} + \frac{1}{x ^{3}} d x

\int_{0}^{4} (x + 4)^{3} d x

\int_{0}^{4} x x + 5 d x

\int_{0}^{1} \frac{x - x}{4} d x

\int_{- 8}^{0} x^{3} d x

What is the integral from sec(θ) to 2 of 3r^2sin(θ) ?
The integral from sec(θ) to 2 of 3r^2sin(θ) is sin(θ)(8-sec^3(θ))