What is the integral from-1 to 1 of sec^3(x) ?

The integral from-1 to 1 of sec^3(x) is sec(1)tan(1)+1/2 (ln(tan(1)+sec(1))-ln(-tan(1)+sec(1)))

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integral from-1 to 1 of sec^3(x)

\int_{- 1}^{1} sec^{3} (x) d x

sec (1) tan (1) + \frac{1}{2} (ln (tan (1) + sec (1)) - ln (- tan (1) + sec (1)))

Decimal Notation

4.10866 \dots

Solution steps

\int_{- 1}^{1} sec^{3} (x) d x

Apply Integral Reduction

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

\int_{- 1}^{1} sec (x) d x = ln (tan (1) + sec (1)) - ln (- tan (1) + sec (1))

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

Simplify

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

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

Compute the boundaries:

sec (1) tan (1)

= sec (1) tan (1) + \frac{1}{2} (ln (tan (1) + sec (1)) - ln (- tan (1) + sec (1)))

f (x) = \frac{( 2 x + 1 )}{x + 2}, at x = 1

y = \frac{3 x - 4}{2 x + 3}

\frac{\partial}{\partial x} (x^{6} y^{6} + 9 x^{4} y)

\frac{d}{d x} ((4 x^{3} + 5 x^{2}) (9 x))

y = 2 x^{2} - 3 x, (2, 2)

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