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

The integral of sin(t)tan(t) is ln|tan(t)+sec(t)|-sin(t)+C

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integral of sin(t)tan(t)

\int sin (t) tan (t) d t

ln ∣ tan (t) + sec (t) ∣ - sin (t) + C

Solution steps

\int sin (t) tan (t) d t

Rewrite using trig identities

= \int \frac{1 - cos ^{2} ( t )}{cos ( t )} d t

Expand

\frac{1 - cos ^{2} ( t )}{cos ( t )} : \frac{1}{cos ( t )} - cos (t)

= \int \frac{1}{cos ( t )} - cos (t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{cos ( t )} d t - \int cos (t) d t

\int \frac{1}{cos ( t )} d t = ln ∣ tan (t) + sec (t) ∣

\int cos (t) d t = sin (t)

= ln ∣ tan (t) + sec (t) ∣ - sin (t)

Add a constant to the solution

= ln ∣ tan (t) + sec (t) ∣ - sin (t) + C

What is the integral of sin(t)tan(t) ?
The integral of sin(t)tan(t) is ln|tan(t)+sec(t)|-sin(t)+C