What is the integral from 0 to 1 of sec^3(x) ?

The integral from 0 to 1 of sec^3(x) is 1/2 sec(1)tan(1)+1/2 ln(tan(1)+sec(1))

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integral from 0 to 1 of sec^3(x)

\int_{0}^{1} sec^{3} (x) d x

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

Decimal

2.05433 \dots

Solution steps

\int_{0}^{1} sec^{3} (x) d x

Apply Integral Reduction

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

\int_{0}^{1} sec (x) d x = ln (tan (1) + sec (1))

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2} sec (1) tan (1)

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

y^{^{'}} = (x + y + 4)^{2}

n = 0 \sum \infty e^{- \frac{7 n}{2}}

\int 7 x^{2} d x

g (x) = 3 x^{4}, (- 2, 48)

\int \frac{( x - 4 )}{( x ^{2} - x + 2 )} d x

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