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

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

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integral from-pi/6 to 0 of 4sec^3(x)

\int_{- \frac{π}{6}}^{0} 4 sec^{3} (x) d x

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

Decimal Notation

2.43194 \dots

Solution steps

\int_{- \frac{π}{6}}^{0} 4 sec^{3} (x) d x

Take the constant out:

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

= 4 \cdot \int_{- \frac{π}{6}}^{0} sec^{3} (x) d x

Apply Integral Reduction

= 4 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{3}

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

Simplify

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

h \to 0 lim (\frac{e ^{h} - 1}{h})

\frac{\partial}{\partial w} (ln (2 w^{3} - 7 y + 3 x y^{2}))

f (x) = cot^{3} (sin (x))

x \to 0 lim (\frac{tan ( 3 x )}{sin ( 7 x )})

\frac{\partial}{\partial y} (z^{3} - x y + yz + y^{3} - 2)

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