What is the integral from 100 to t of sec(2x-7) ?

The integral from 100 to t of sec(2x-7) is 1/2 (ln|tan(-7+2t)+sec(-7+2t)|-ln(-tan(193)-sec(193)))

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integral from 100 to t of sec(2x-7)

\int_{100}^{t} sec (2 x - 7) d x

\frac{1}{2} (ln ∣ tan (- 7 + 2 t) + sec (- 7 + 2 t) ∣ - ln (- tan (193) - sec (193)))

Solution steps

\int_{100}^{t} sec (2 x - 7) d x

Apply u-substitution

: \int_{193}^{- 7 + 2 t} sec (u) \frac{1}{2} d u

= \int_{193}^{- 7 + 2 t} sec (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_{193}^{- 7 + 2 t} sec (u) d u

Use the common integral:

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

= \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{193}^{- 7 + 2 t}

Compute the boundaries:

ln ∣ tan (- 7 + 2 t) + sec (- 7 + 2 t) ∣ - ln (- tan (193) - sec (193))

= \frac{1}{2} (ln ∣ tan (- 7 + 2 t) + sec (- 7 + 2 t) ∣ - ln (- tan (193) - sec (193)))

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What is the integral from 100 to t of sec(2x-7) ?
The integral from 100 to t of sec(2x-7) is 1/2 (ln|tan(-7+2t)+sec(-7+2t)|-ln(-tan(193)-sec(193)))