What is the integral from 0 to x of sqrt(cos(2t)) ?

The integral from 0 to x of sqrt(cos(2t)) is \E(2\vert (2x)/2)-\E(0*2\vert)

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integral from 0 to x of sqrt(cos(2t))

\int_{0}^{x} cos (2 t) d t

E (2∣ \frac{2 x}{2}) - E (0 \cdot 2∣)

Solution steps

\int_{0}^{x} cos (2 t) d t

Apply u-substitution

: \int_{0}^{2 x} cos (u) \frac{1}{2} d u

= \int_{0}^{2 x} cos (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}^{2 x} cos (u) d u

Use the nonelementary integral:

\int cos (x) d x = 2 E (\frac{x}{2} ∣ 2)

= \frac{1}{2} [2 E (\frac{u}{2} ∣ 2)]_{0}^{2 x}

Compute the boundaries:

2 E (\frac{2 x}{2} ∣ 2) - 2 E (0 \cdot 2∣)

= \frac{1}{2} (2 E (\frac{2 x}{2} ∣ 2) - 2 E (0 \cdot 2∣))

Simplify

= E (2∣ \frac{2 x}{2}) - E (0 \cdot 2∣)

\int 4 \frac{ln ( x )}{x} d x

\frac{4 x}{9 - tan ( x )}

x \to e lim (ln (x))

\int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

y^{^{'}} = \frac{2 x y}{x ^{2} + 1}, y (1) = 6

What is the integral from 0 to x of sqrt(cos(2t)) ?
The integral from 0 to x of sqrt(cos(2t)) is \E(2\vert (2x)/2)-\E(0*2\vert)