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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Apply u-substitution

: \int_{0}^{\frac{π}{4}} sec^{2} (u) \frac{1}{2} d u

= \int_{0}^{\frac{π}{4}} sec^{2} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= \frac{1}{2} [tan (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

1

= \frac{1}{2} \cdot 1

Simplify

= \frac{1}{2}

\int_{0}^{\frac{π}{6}} 1

x \to 0 - lim (x^{s i n (x)})

y = (1 + x)^{3}

\frac{d}{d t} (sec^{2} (t))

\frac{x ^{2} - 100}{x + 10}

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