What is the integral of tan^3(7x)sec(7x) ?

The integral of tan^3(7x)sec(7x) is -1/7 sec(7x)+(sec^3(7x))/(21)+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(7x)sec(7x)

\int tan^{3} (7 x) sec (7 x) d x

- \frac{1}{7} sec (7 x) + \frac{sec ^{3} ( 7 x )}{21} + C

Solution steps

\int tan^{3} (7 x) sec (7 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (7 x)) tan (7 x) sec (7 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{7} d u

= \int \frac{- 1 + u ^{2}}{7} d u

Expand

\frac{- 1 + u ^{2}}{7} : - \frac{1}{7} + \frac{u ^{2}}{7}

= \int - \frac{1}{7} + \frac{u ^{2}}{7} 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}{7} d u + \int \frac{u ^{2}}{7} d u

\int \frac{1}{7} d u = \frac{1}{7} u

\int \frac{u ^{2}}{7} d u = \frac{u ^{3}}{21}

= - \frac{1}{7} u + \frac{u ^{3}}{21}

Substitute back

u = sec (7 x)

= - \frac{1}{7} sec (7 x) + \frac{sec ^{3} ( 7 x )}{21}

Add a constant to the solution

= - \frac{1}{7} sec (7 x) + \frac{sec ^{3} ( 7 x )}{21} + C

- 2 x^{3}

\frac{d y}{d x} + y = 1

\int 2 x^{7} d x

\frac{d ^{2}}{d x ^{2}} ((ln (x))^{2})

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

What is the integral of tan^3(7x)sec(7x) ?
The integral of tan^3(7x)sec(7x) is -1/7 sec(7x)+(sec^3(7x))/(21)+C