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

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

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integral of sec(t)tan(t)

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

sec (t) + C

Solution steps

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

Apply u-substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = sec (t)

= 1 \cdot sec (t)

Simplify

= sec (t)

Add a constant to the solution

= sec (t) + C

g (x) = x^{2} (1 - 4 x)

\frac{d}{d x} (e^{x} (1 - x^{2}))

f (x) = \frac{8}{x}

x y^{^{'}} = 3 x^{3} y^{\frac{1}{2}} - 2 y

\frac{s}{( s + 1 ) ^{2} + 1}