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

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

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

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

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

Decimal

3.64791 \dots

Solution steps

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

Apply Integral Reduction

= 6 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{3}

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

sin (6 (t - \frac{π}{6}))

y = \frac{t}{( t - 4 ) ^{2}}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = sin (x)

n = 0 \sum \infty \frac{1}{( 4 n + 3 ) ^{2}}

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

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