What is the integral from 0 to pi/2 of sec(x)tan(x) ?

The integral from 0 to pi/2 of sec(x)tan(x) is diverges

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integral from 0 to pi/2 of sec(x)tan(x)

\int_{0}^{\frac{π}{2}} sec (x) tan (x) d x

diverges

Solution steps

\int_{0}^{\frac{π}{2}} sec (x) tan (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{sin ( x )}{cos ^{2} ( x )} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{u ^{2}} d u

= \int_{1}^{0} - \frac{1}{u ^{2}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{1}{u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{0}^{1} \frac{1}{u ^{2}} d u)

Apply the Power Rule

: [- \frac{1}{u}]_{0}^{1}

= - (- [- \frac{1}{u}]_{0}^{1})

Simplify

= [- \frac{1}{u}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to pi/2 of sec(x)tan(x) ?
The integral from 0 to pi/2 of sec(x)tan(x) is diverges