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

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

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

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

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

Decimal Notation

3.64791 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Reduction

= 6 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{3}

= 6 (\frac{1}{3} + \frac{1}{4} ln (3))

\frac{d}{d x} (- sin (x) cos (x))

\frac{d}{d x} (\frac{2 x - 3}{5 x + 4})

\int (x + 6) (x - 4) d x

\int \frac{5}{2 x - 1} d x

\int_{5}^{8} (x - 7)^{2} d x

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