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

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

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integral from-pi/4 to 0 of 2sec^3(x)

\int_{- \frac{π}{4}}^{0} 2 sec^{3} (x) d x

2 - ln (2 - 1)

Decimal Notation

2.29558 \dots

Solution steps

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

Apply Integral Reduction

= 2 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

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

Simplify

= 2 - ln (2 - 1)

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