What is the integral of tan^2(6x)sec^3(6x) ?

The integral of tan^2(6x)sec^3(6x) is 1/48 (-sec(6x)tan(6x)-ln|tan(6x)+sec(6x)|+2sec^4(6x)sin(6x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^2(6x)sec^3(6x)

\int tan^{2} (6 x) sec^{3} (6 x) d x

\frac{1}{48} (- sec (6 x) tan (6 x) - ln ∣ tan (6 x) + sec (6 x) ∣ + 2 sec^{4} (6 x) sin (6 x)) + C

Solution steps

\int tan^{2} (6 x) sec^{3} (6 x) d x

Apply u-substitution

: \int tan^{2} (u) sec^{3} (u) \frac{1}{6} d u

= \int tan^{2} (u) sec^{3} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int tan^{2} (u) sec^{3} (u) d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int (- 1 + sec^{2} (u)) sec^{3} (u) d u

Expand

(- 1 + sec^{2} (u)) sec^{3} (u) : - sec^{3} (u) + sec^{5} (u)

= \frac{1}{6} \cdot \int - sec^{3} (u) + sec^{5} (u) d u

Apply the Sum Rule:

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

= \frac{1}{6} (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

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

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= \frac{1}{6} (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Substitute back

u = 6 x

= \frac{1}{6} (- (\frac{1}{2} sec (6 x) tan (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣) + \frac{sec ^{4} ( 6 x ) sin ( 6 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (6 x) tan (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣))

Simplify

\frac{1}{6} (- (\frac{1}{2} sec (6 x) tan (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣) + \frac{sec ^{4} ( 6 x ) sin ( 6 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (6 x) tan (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣)) : \frac{1}{48} (- sec (6 x) tan (6 x) - ln ∣ tan (6 x) + sec (6 x) ∣ + 2 sec^{4} (6 x) sin (6 x))

= \frac{1}{48} (- sec (6 x) tan (6 x) - ln ∣ tan (6 x) + sec (6 x) ∣ + 2 sec^{4} (6 x) sin (6 x))

Add a constant to the solution

= \frac{1}{48} (- sec (6 x) tan (6 x) - ln ∣ tan (6 x) + sec (6 x) ∣ + 2 sec^{4} (6 x) sin (6 x)) + C

\int 6 - 2 x d x

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

\int \frac{3 x}{e ^{2 x}} d x

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

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

What is the integral of tan^2(6x)sec^3(6x) ?
The integral of tan^2(6x)sec^3(6x) is 1/48 (-sec(6x)tan(6x)-ln|tan(6x)+sec(6x)|+2sec^4(6x)sin(6x))+C