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

The integral of tan^3(3x)sec(3x) is -1/3 sec(3x)+(sec^3(3x))/9+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{3} sec (3 x) + \frac{sec ^{3} ( 3 x )}{9} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{3} d u

= \int \frac{- 1 + u ^{2}}{3} d u

Expand

\frac{- 1 + u ^{2}}{3} : - \frac{1}{3} + \frac{u ^{2}}{3}

= \int - \frac{1}{3} + \frac{u ^{2}}{3} 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

= - \int \frac{1}{3} d u + \int \frac{u ^{2}}{3} d u

\int \frac{1}{3} d u = \frac{1}{3} u

\int \frac{u ^{2}}{3} d u = \frac{u ^{3}}{9}

= - \frac{1}{3} u + \frac{u ^{3}}{9}

Substitute back

u = sec (3 x)

= - \frac{1}{3} sec (3 x) + \frac{sec ^{3} ( 3 x )}{9}

Add a constant to the solution

= - \frac{1}{3} sec (3 x) + \frac{sec ^{3} ( 3 x )}{9} + C

\int \frac{1}{( k - 1 ) ^{4}}

y = \frac{3 x ^{2} + 2}{x + 1}

\int \frac{1}{x + 1 + x} d x

\frac{d ^{3}}{d x ^{3}} (4 x - 3)

\int \frac{2 x + 6}{x ^{2} + 1} d x

What is the integral of tan^3(3x)sec(3x) ?
The integral of tan^3(3x)sec(3x) is -1/3 sec(3x)+(sec^3(3x))/9+C