What is the integral from 0 to pi/6 of 2sec^3(x) ?

The integral from 0 to pi/6 of 2sec^3(x) is 2(1/3+1/4 ln(3))

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integral from 0 to pi/6 of 2sec^3(x)

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

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

Decimal

1.21597 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Reduction

= 2 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{3}

= 2 (\frac{1}{3} + \frac{1}{4} ln (3))

\int 1 - e^{- x} d x

\int_{1}^{7} \frac{x + 1}{x ^{2} + 2 x - 1} d x

y = \frac{π ^{2} x ^{6} + 9000}{x ^{2}}

\int 2 tan^{4} (x se) c^{6} x d x

f (x) = e^{x}, at x = 7

What is the integral from 0 to pi/6 of 2sec^3(x) ?
The integral from 0 to pi/6 of 2sec^3(x) is 2(1/3+1/4 ln(3))