What is the integral of sec(t)(8sec(t)+9tan(t)) ?

The integral of sec(t)(8sec(t)+9tan(t)) is 8tan(t)+9sec(t)+C

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integral of sec(t)(8sec(t)+9tan(t))

\int sec (t) (8 sec (t) + 9 tan (t)) d t

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

Solution steps

\int sec (t) (8 sec (t) + 9 tan (t)) d t

Rewrite using trig identities

= \int \frac{8 + 9 sin ( t )}{cos ^{2} ( t )} d t

Expand

\frac{8 + 9 sin ( t )}{cos ^{2} ( t )} : \frac{8}{cos ^{2} ( t )} + \frac{9 sin ( t )}{cos ^{2} ( t )}

= \int \frac{8}{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{8}{cos ^{2} ( t )} d t + \int \frac{9 sin ( t )}{cos ^{2} ( t )} d t

\int \frac{8}{cos ^{2} ( t )} d t = 8 tan (t)

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

= 8 tan (t) + 9 sec (t)

Add a constant to the solution

= 8 tan (t) + 9 sec (t) + C

x \to 1 lim (2 x^{2} + 3 x + 1)

\int_{5}^{9} \frac{x}{x ^{2} - 10 x + 41} d x

f (x) = 2 t^{- \frac{1}{4}}

sin (5 ln (x))

f (x) = \frac{2 x}{x ^{2} + 1}, at x = 0

What is the integral of sec(t)(8sec(t)+9tan(t)) ?
The integral of sec(t)(8sec(t)+9tan(t)) is 8tan(t)+9sec(t)+C