What is the integral of sec(t)(4sec(t)+9tan(t)) ?

The integral of sec(t)(4sec(t)+9tan(t)) is 4tan(t)+9sec(t)+C

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integral of sec(t)(4sec(t)+9tan(t))

\int sec (t) (4 sec (t) + 9 tan (t)) d t

4 tan (t) + 9 sec (t) + C

Solution steps

\int sec (t) (4 sec (t) + 9 tan (t)) d t

Rewrite using trig identities

= \int \frac{4 + 9 sin ( t )}{cos ^{2} ( t )} d t

Expand

\frac{4 + 9 sin ( t )}{cos ^{2} ( t )} : \frac{4}{cos ^{2} ( t )} + \frac{9 sin ( t )}{cos ^{2} ( t )}

= \int \frac{4}{cos ^{2} ( t )} + \frac{9 sin ( t )}{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{4}{cos ^{2} ( t )} d t + \int \frac{9 sin ( t )}{cos ^{2} ( t )} d t

\int \frac{4}{cos ^{2} ( t )} d t = 4 tan (t)

\int \frac{9 sin ( t )}{cos ^{2} ( t )} d t = 9 sec (t)

= 4 tan (t) + 9 sec (t)

Add a constant to the solution

= 4 tan (t) + 9 sec (t) + C

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What is the integral of sec(t)(4sec(t)+9tan(t)) ?
The integral of sec(t)(4sec(t)+9tan(t)) is 4tan(t)+9sec(t)+C