What is the integral of sec(t)(4sec(t)+7tan(t)) ?

The integral of sec(t)(4sec(t)+7tan(t)) is 4tan(t)+7sec(t)+C

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integral of sec(t)(4sec(t)+7tan(t))

\int sec (t) (4 sec (t) + 7 tan (t)) d t

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

Solution steps

\int sec (t) (4 sec (t) + 7 tan (t)) d t

Rewrite using trig identities

= \int \frac{4 + 7 sin ( t )}{cos ^{2} ( t )} d t

Expand

\frac{4 + 7 sin ( t )}{cos ^{2} ( t )} : \frac{4}{cos ^{2} ( t )} + \frac{7 sin ( t )}{cos ^{2} ( t )}

= \int \frac{4}{cos ^{2} ( t )} + \frac{7 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{7 sin ( t )}{cos ^{2} ( t )} d t

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

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

= 4 tan (t) + 7 sec (t)

Add a constant to the solution

= 4 tan (t) + 7 sec (t) + C

\frac{d}{d y} (y e^{x})

x \to - \infty lim (x^{- 2})

\frac{d}{d x} (\frac{8 x}{9 - cot ( x )})

x^{2} y^{^{''}} + 7 x y^{^{'}} + 8 y = 0, y (1) = 0, y^{^{'}} (1) = - 5

\frac{\partial}{\partial y} (ln (2 x + 3 y))

What is the integral of sec(t)(4sec(t)+7tan(t)) ?
The integral of sec(t)(4sec(t)+7tan(t)) is 4tan(t)+7sec(t)+C