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

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

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

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

2 (- ln (2 - 1) + 2)

Decimal

4.59117 \dots

Solution steps

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

Apply Integral Reduction

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

\int_{- \frac{π}{4}}^{0} sec (x) d x = - ln (2 - 1)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

= 4 (\frac{1}{2} - \frac{1}{2} ln (2 - 1))

Simplify

= 2 (- ln (2 - 1) + 2)

\frac{\partial}{\partial x} (2 x^{2} + Y (x) (x)^{2})

\int \frac{2 x}{x} d x

\int \frac{x + 1}{x ^{2} - 3 x + 2} d x

x - 2 x^{2}

\frac{d ^{3}}{d x ^{3}} (e^{5 x - 4})

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