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

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

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

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

Solution steps

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

\frac{\partial}{\partial x} (x + x y + y)

\frac{2 x}{x ^{2} + 1}, \frac{x}{13}, [- 5, 5]

3 sin (x), 4 cos (x)

x + 5 x y - 2 x^{3} = 2

y^{^{''}} = x sin (x) + x cos (x)

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