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

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

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

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

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

Decimal Notation

2.43194 \dots

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{3}

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

Simplify

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

\frac{d ^{3}}{d x ^{3}} (\frac{1}{x - 1})

\int 16 tan^{4} (x) sec^{6} (x) d x

\int \frac{1}{sin ^{3} ( x )} d x

f (x) = 38 - x^{2}, (- 6, 2)

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