What is the integral of 3tan^2(x)sec(x) ?

The integral of 3tan^2(x)sec(x) is 3(1/2 sec(x)tan(x)-1/2 ln|tan(x)+sec(x)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3tan^2(x)sec(x)

\int 3 tan^{2} (x) sec (x) d x

3 (\frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

Solution steps

\int 3 tan^{2} (x) sec (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int tan^{2} (x) sec (x) d x

Rewrite using trig identities

= 3 \cdot \int (- 1 + sec^{2} (x)) sec (x) d x

Expand

(- 1 + sec^{2} (x)) sec (x) : - sec (x) + sec^{3} (x)

= 3 \cdot \int - sec (x) + sec^{3} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (- \int sec (x) d x + \int sec^{3} (x) d x)

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= 3 (- ln ∣ tan (x) + sec (x) ∣ + \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Simplify

- ln ∣ tan (x) + sec (x) ∣ + \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣ : \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= 3 (\frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Add a constant to the solution

= 3 (\frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

a x^{2} + b x + c

\frac{d}{d x} (- 4 cos (4 x))

\frac{\partial}{\partial y} (sin (x) + π cos (y))

2 v^{'} t (x) = v

\int \frac{( 1 - y ) e ^{y}}{y ^{2}} d y

What is the integral of 3tan^2(x)sec(x) ?
The integral of 3tan^2(x)sec(x) is 3(1/2 sec(x)tan(x)-1/2 ln|tan(x)+sec(x)|)+C