What is the integral from-pi/4 to 0 of 8sec^3(x) ?

The integral from-pi/4 to 0 of 8sec^3(x) is 4(-ln(sqrt(2)-1)+sqrt(2))

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integral from-pi/4 to 0 of 8sec^3(x)

\int_{- \frac{π}{4}}^{0} 8 sec^{3} (x) d x

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

Decimal Notation

9.18234 \dots

Solution steps

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

Apply Integral Reduction

= 8 ([\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)

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} (- ln (2 - 1)))

Simplify

8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} (- ln (2 - 1))) : 8 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{4}}^{0} - \frac{1}{2} ln (2 - 1))

= 8 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{4}}^{0} - \frac{1}{2} ln (2 - 1))

Compute the boundaries:

\frac{1}{2}

= 8 (\frac{1}{2} - \frac{1}{2} ln (2 - 1))

Simplify

= 4 (- ln (2 - 1) + 2)

\frac{\partial}{\partial x} (3 x^{2} + 2 x y + y^{2} - 11)

15 t^{3}

f (x) = sin (x cos (x))

t \frac{d y}{d t} + 2 y = sin (t)

\int_{0}^{3} (e^{x}) d x

What is the integral from-pi/4 to 0 of 8sec^3(x) ?
The integral from-pi/4 to 0 of 8sec^3(x) is 4(-ln(sqrt(2)-1)+sqrt(2))