What is the integral from 0 to 2 of 2sec((pix)/6) ?

The integral from 0 to 2 of 2sec((pix)/6) is (12)/pi ln(2+sqrt(3))

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integral from 0 to 2 of 2sec((pix)/6)

\int_{0}^{2} 2 sec (\frac{π x}{6}) d x

\frac{12}{π} ln (2 + 3)

Solution steps

\int_{0}^{2} 2 sec (\frac{π x}{6}) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} sec (\frac{π x}{6}) d x

Apply u-substitution

: \int_{0}^{\frac{π}{3}} sec (u) \frac{6}{π} d u

= 2 \cdot \int_{0}^{\frac{π}{3}} sec (u) \frac{6}{π} d u

Take the constant out:

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

= 2 \cdot \frac{6}{π} \cdot \int_{0}^{\frac{π}{3}} sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 2 \cdot \frac{6}{π} [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{3}}

Simplify

2 \cdot \frac{6}{π} [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{3}} : \frac{12}{π} [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{3}}

= \frac{12}{π} [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{3}}

Compute the boundaries:

ln (2 + 3)

= \frac{12}{π} ln (2 + 3)

2^{x}, 1

\int \frac{1}{θ ^{2}} sin (\frac{1}{θ}) cos (\frac{1}{θ}) d θ

\frac{d}{d t} (e^{5 t})

y = e^{9 x}

\frac{d}{d t} (e^{t} cos (2 t))

What is the integral from 0 to 2 of 2sec((pix)/6) ?
The integral from 0 to 2 of 2sec((pix)/6) is (12)/pi ln(2+sqrt(3))