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

The integral from 0 to pi/8 of x(sec(2x))^2 is 1/2 (pi/8-1/4 ln(2))

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

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

\frac{1}{2} (\frac{π}{8} - \frac{1}{4} ln (2))

Decimal Notation

0.10970 \dots

Solution steps

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

Simplify

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

Apply Integration By Parts

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

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

\int \frac{1}{2} tan (2 x) d x = - \frac{1}{4} ln ∣ cos (2 x) ∣

= [\frac{1}{2} (x tan (2 x) - 2 (- \frac{1}{4} ln ∣ cos (2 x) ∣))]_{0}^{\frac{π}{8}}

Simplify

[\frac{1}{2} (x tan (2 x) - 2 (- \frac{1}{4} ln ∣ cos (2 x) ∣))]_{0}^{\frac{π}{8}} : [\frac{1}{2} (x tan (2 x) + \frac{1}{2} ln ∣ cos (2 x) ∣)]_{0}^{\frac{π}{8}}

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

Compute the boundaries:

\frac{1}{2} (\frac{π}{8} - \frac{1}{4} ln (2))

= \frac{1}{2} (\frac{π}{8} - \frac{1}{4} ln (2))

\frac{d}{d x} (\frac{2 x ^{2} - 1}{x ^{3 - x}})

\frac{d}{d x} (- 11 (3)^{x - 4})

\int arctan (9 t) d t

\frac{d y}{d x} + y^{7} x + 6 y = 0

(- 1, 1), (2, - 11)

What is the integral from 0 to pi/8 of x(sec(2x))^2 ?
The integral from 0 to pi/8 of x(sec(2x))^2 is 1/2 (pi/8-1/4 ln(2))