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

The integral from-pi/6 to 0 of 2sec^3(x) is 2(1/3+1/4 ln(3))

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integral from-pi/6 to 0 of 2sec^3(x)

\int_{- \frac{π}{6}}^{0} 2 sec^{3} (x) d x

2 (\frac{1}{3} + \frac{1}{4} ln (3))

Decimal

1.21597 \dots

Solution steps

\int_{- \frac{π}{6}}^{0} 2 sec^{3} (x) d x

Take the constant out:

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

= 2 \cdot \int_{- \frac{π}{6}}^{0} sec^{3} (x) d x

Apply Integral Reduction

= 2 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \int_{- \frac{π}{6}}^{0} sec (x) d x)

\int_{- \frac{π}{6}}^{0} sec (x) d x = \frac{1}{2} ln (3)

= 2 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \frac{1}{2} ln (3))

Simplify

2 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \frac{1}{2} ln (3)) : 2 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{6}}^{0} + \frac{1}{4} ln (3))

= 2 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{6}}^{0} + \frac{1}{4} ln (3))

Compute the boundaries:

\frac{1}{3}

= 2 (\frac{1}{3} + \frac{1}{4} ln (3))

\frac{\partial}{\partial y} (x^{a} y^{b} z^{c})

f (x) = \frac{3}{x ^{2} + 3}

\frac{e ^{x} - e ^{- x}}{2}

\frac{d}{d x} (ln (e^{3 x^{2}} \cdot 4^{5 x}))

\frac{\partial}{\partial x} ((1 + x y)^{\frac{3}{2}})

What is the integral from-pi/6 to 0 of 2sec^3(x) ?
The integral from-pi/6 to 0 of 2sec^3(x) is 2(1/3+1/4 ln(3))