What is the integral from 0 to pi/3 of 8sec^3(x) ?

The integral from 0 to pi/3 of 8sec^3(x) is 8(sqrt(3)+1/2 ln(2+sqrt(3)))

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integral from 0 to pi/3 of 8sec^3(x)

\int_{0}^{\frac{π}{3}} 8 sec^{3} (x) d x

8 (3 + \frac{1}{2} ln (2 + 3))

Decimal Notation

19.12423 \dots

Solution steps

\int_{0}^{\frac{π}{3}} 8 sec^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{\frac{π}{3}} sec^{3} (x) d x

Apply Integral Reduction

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{3}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} sec (x) d x)

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

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3))

Simplify

8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3)) : 8 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3))

= 8 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3))

Compute the boundaries:

3

= 8 (3 + \frac{1}{2} ln (2 + 3))

f (x) = 7 x + 2, (2, 4)

x y^{^{'}} + y = e^{x}, y (1) = 1

(3, 6), (5, - 8)

\frac{\partial}{\partial x} (ln (x + t^{2}))

x^{2} - 10 x + 4

What is the integral from 0 to pi/3 of 8sec^3(x) ?
The integral from 0 to pi/3 of 8sec^3(x) is 8(sqrt(3)+1/2 ln(2+sqrt(3)))