What is the integral of 1/(cos(x-1)) ?

The integral of 1/(cos(x-1)) is ln|tan(x-1)+sec(x-1)|+C

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integral of 1/(cos(x-1))

\int \frac{1}{cos ( x - 1 )} d x

ln ∣ tan (x - 1) + sec (x - 1) ∣ + C

Solution steps

\int \frac{1}{cos ( x - 1 )} d x

Rewrite using trig identities

= \int sec (x - 1) d x

Apply u-substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

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

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = x - 1

= ln ∣ tan (x - 1) + sec (x - 1) ∣

Add a constant to the solution

= ln ∣ tan (x - 1) + sec (x - 1) ∣ + C

f (x) = x^{3} - 2 x

x \to 2 lim ([6 f (x) + g (x)])

4 x + 10, x^{2} + 6 x + 6, - 2, 2

(x^{2} + 16) \frac{d y}{d x} + x y = 5 x

lo g_{8} (5 x^{3} + 3 x)

What is the integral of 1/(cos(x-1)) ?
The integral of 1/(cos(x-1)) is ln|tan(x-1)+sec(x-1)|+C