What is the integral of sec(t-1)tan(t-1) ?

The integral of sec(t-1)tan(t-1) is sec(t-1)+C

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integral of sec(t-1)tan(t-1)

\int sec (t - 1) tan (t - 1) d t

sec (t - 1) + C

Solution steps

\int sec (t - 1) tan (t - 1) d t

Apply u-substitution

: \int sec (u) tan (u) d u

= \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot sec (t - 1)

Simplify

= sec (t - 1)

Add a constant to the solution

= sec (t - 1) + C

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What is the integral of sec(t-1)tan(t-1) ?
The integral of sec(t-1)tan(t-1) is sec(t-1)+C