What is the integral from 0 to pi/6 of (9sec^2(x)) ?

The integral from 0 to pi/6 of (9sec^2(x)) is 3sqrt(3)

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integral from 0 to pi/6 of (9sec^2(x))

\int_{0}^{\frac{π}{6}} (9 sec^{2} (x)) d x

33

Solution steps

\int_{0}^{\frac{π}{6}} 9 sec^{2} (x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{\frac{π}{6}} sec^{2} (x) d x

Use the common integral:

\int sec^{2} (x) d x = tan (x)

= 9 [tan (x)]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{1}{3}

= 9 \cdot \frac{1}{3}

Simplify

= 33

\int \frac{1}{7 - 3 x} d x

y^{^{'}} = 9 x + y

(2 e^{y} - x) (\frac{d y}{d x}) = 1

f (x) = - 7 e^{- 3 x} - 6 e^{- 3 x} x

x \to 3 lim (\frac{x ^{2} - 8}{x - 3})

What is the integral from 0 to pi/6 of (9sec^2(x)) ?
The integral from 0 to pi/6 of (9sec^2(x)) is 3sqrt(3)