What is the integral from 0 to pi/8 of sec(2x)tan(2x) ?

The integral from 0 to pi/8 of sec(2x)tan(2x) is 1/(sqrt(2))-1/2

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integral from 0 to pi/8 of sec(2x)tan(2x)

\int_{0}^{\frac{π}{8}} sec (2 x) tan (2 x) d x

\frac{1}{2} - \frac{1}{2}

Decimal

0.20710 \dots

Solution steps

\int_{0}^{\frac{π}{8}} sec (2 x) tan (2 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{8}} \frac{sin ( 2 x )}{cos ^{2} ( 2 x )} d x

Apply Integration By Parts

: [\frac{1}{2} (sin (2 x) tan (2 x) - 2 \cdot \int sin (2 x) d x)]_{0}^{\frac{π}{8}}

= [\frac{1}{2} (sin (2 x) tan (2 x) - 2 \cdot \int sin (2 x) d x)]_{0}^{\frac{π}{8}}

\int sin (2 x) d x = - \frac{1}{2} cos (2 x)

= [\frac{1}{2} (sin (2 x) tan (2 x) - 2 (- \frac{1}{2} cos (2 x)))]_{0}^{\frac{π}{8}}

Simplify

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

= [\frac{1}{2} (sin (2 x) tan (2 x) + cos (2 x))]_{0}^{\frac{π}{8}}

Compute the boundaries:

\frac{1}{2} - \frac{1}{2}

= \frac{1}{2} - \frac{1}{2}

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What is the integral from 0 to pi/8 of sec(2x)tan(2x) ?
The integral from 0 to pi/8 of sec(2x)tan(2x) is 1/(sqrt(2))-1/2