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

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

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

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

6 (3 - \frac{1}{2} ln (2 - 3))

Decimal

14.34317 \dots

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

3

= 6 (3 - \frac{1}{2} ln (2 - 3))

\int x^{2} y e^{x y} d x

(y - x 3) d x + (x + y 3) d y = 0

f^{^{'}} (x) = x e^{x^{2}}

\frac{\partial}{\partial x} (x + 6 y)

\int 4 e^{7} d x

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