What is the integral of sec(t)[tan(t)+sec(t)] ?

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

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integral of sec(t)[tan(t)+sec(t)]

\int sec (t) [tan (t) + sec (t)] d t

sec (t) + tan (t) + C

Solution steps

\int sec (t) [tan (t) + sec (t)] d t

Rewrite using trig identities

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

Expand

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

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

\int \frac{sin ( t )}{cos ^{2} ( t )} d t = sec (t)

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

= sec (t) + tan (t)

Add a constant to the solution

= sec (t) + tan (t) + C

What is the integral of sec(t)[tan(t)+sec(t)] ?
The integral of sec(t)[tan(t)+sec(t)] is sec(t)+tan(t)+C