What is the integral of tan^5(x)sec(x) ?

The integral of tan^5(x)sec(x) is sec(x)-(2sec^3(x))/3+(sec^5(x))/5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^5(x)sec(x)

\int tan^{5} (x) sec (x) d x

sec (x) - \frac{2 sec ^{3} ( x )}{3} + \frac{sec ^{5} ( x )}{5} + C

Solution steps

\int tan^{5} (x) sec (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int (- 1 + u^{2})^{2} d u

= \int (- 1 + u^{2})^{2} d u

Expand

(- 1 + u^{2})^{2} : 1 - 2 u^{2} + u^{4}

= \int 1 - 2 u^{2} + u^{4} 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 1 d u - \int 2 u^{2} d u + \int u^{4} d u

\int 1 d u = u

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

\int u^{4} d u = \frac{u ^{5}}{5}

= u - \frac{2 u ^{3}}{3} + \frac{u ^{5}}{5}

Substitute back

u = sec (x)

= sec (x) - \frac{2 sec ^{3} ( x )}{3} + \frac{sec ^{5} ( x )}{5}

Add a constant to the solution

= sec (x) - \frac{2 sec ^{3} ( x )}{3} + \frac{sec ^{5} ( x )}{5} + C

\int \frac{x ^{2} - 3 x - 5}{x ^{2} - 5 x} d x

y \to 0 lim (\frac{7}{y ^{2} + y} - \frac{7}{y})

5 e^{- x}

f^{^{'}} (x) = sin (x^{2})

\frac{d}{d x} ((sin (x))^{l n (x)})

What is the integral of tan^5(x)sec(x) ?
The integral of tan^5(x)sec(x) is sec(x)-(2sec^3(x))/3+(sec^5(x))/5+C