What is the integral of sec(t)(4sec(t)+3tan(t)) ?

The integral of sec(t)(4sec(t)+3tan(t)) is 4tan(t)+3sec(t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sec(t)(4sec(t)+3tan(t))

\int sec (t) (4 sec (t) + 3 tan (t)) d t

4 tan (t) + 3 sec (t) + C

Solution steps

\int sec (t) (4 sec (t) + 3 tan (t)) d t

Rewrite using trig identities

= \int \frac{4 + 3 sin ( t )}{cos ^{2} ( t )} d t

Expand

\frac{4 + 3 sin ( t )}{cos ^{2} ( t )} : \frac{4}{cos ^{2} ( t )} + \frac{3 sin ( t )}{cos ^{2} ( t )}

= \int \frac{4}{cos ^{2} ( t )} + \frac{3 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{3 sin ( t )}{cos ^{2} ( t )} d t

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

\int \frac{3 sin ( t )}{cos ^{2} ( t )} d t = 3 sec (t)

= 4 tan (t) + 3 sec (t)

Add a constant to the solution

= 4 tan (t) + 3 sec (t) + C

\frac{\partial}{\partial y} (y ln (x^{2} + y^{3} + 5))

\int d x d x

\frac{d}{d x} (\frac{( x ^{2} + 4 x + 3 )}{x})

\int (\frac{1}{x ^{2}} + x) d x

n = 1 \sum \infty 8 (- 1)^{n} n x^{n}

What is the integral of sec(t)(4sec(t)+3tan(t)) ?
The integral of sec(t)(4sec(t)+3tan(t)) is 4tan(t)+3sec(t)+C