What is the integral from-pi/6 to 0 of 8sec^3(x) ?

The integral from-pi/6 to 0 of 8sec^3(x) is 8(1/3+1/4 ln(3))

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integral from-pi/6 to 0 of 8sec^3(x)

\int_{- \frac{π}{6}}^{0} 8 sec^{3} (x) d x

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

Decimal Notation

4.86389 \dots

Solution steps

\int_{- \frac{π}{6}}^{0} 8 sec^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int_{- \frac{π}{6}}^{0} sec^{3} (x) d x

Apply Integral Reduction

= 8 ([\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)

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \frac{1}{2} ln (3))

Simplify

8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \frac{1}{2} ln (3)) : 8 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{6}}^{0} + \frac{1}{4} ln (3))

= 8 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{6}}^{0} + \frac{1}{4} ln (3))

Compute the boundaries:

\frac{1}{3}

= 8 (\frac{1}{3} + \frac{1}{4} ln (3))

\frac{\partial}{\partial x} (ln (\frac{k}{x}))

\int e^{5 x} cos (6 x) d x

\frac{d}{d x} ((ln (x^{2}))^{- 3})

\frac{d}{d x} (2 sin (π) x + 3 cos (π) x)

f (x) = \frac{7}{1 - x}

What is the integral from-pi/6 to 0 of 8sec^3(x) ?
The integral from-pi/6 to 0 of 8sec^3(x) is 8(1/3+1/4 ln(3))