What is the integral of sec(t)(7sec(t)+8tan(t)) ?

The integral of sec(t)(7sec(t)+8tan(t)) is 7tan(t)+8sec(t)+C

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integral of sec(t)(7sec(t)+8tan(t))

\int sec (t) (7 sec (t) + 8 tan (t)) d t

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

Solution steps

\int sec (t) (7 sec (t) + 8 tan (t)) d t

Rewrite using trig identities

= \int \frac{7 + 8 sin ( t )}{cos ^{2} ( t )} d t

Expand

\frac{7 + 8 sin ( t )}{cos ^{2} ( t )} : \frac{7}{cos ^{2} ( t )} + \frac{8 sin ( t )}{cos ^{2} ( t )}

= \int \frac{7}{cos ^{2} ( t )} + \frac{8 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{7}{cos ^{2} ( t )} d t + \int \frac{8 sin ( t )}{cos ^{2} ( t )} d t

\int \frac{7}{cos ^{2} ( t )} d t = 7 tan (t)

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

= 7 tan (t) + 8 sec (t)

Add a constant to the solution

= 7 tan (t) + 8 sec (t) + C

\int 3 t - 7 d t

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\int 3 x^{3} x^{2} + 1 d x

What is the integral of sec(t)(7sec(t)+8tan(t)) ?
The integral of sec(t)(7sec(t)+8tan(t)) is 7tan(t)+8sec(t)+C