What is the integral of sqrt(tan(x))(sec(x))^4 ?

The integral of sqrt(tan(x))(sec(x))^4 is 2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(tan(x))(sec(x))^4

\int tan (x) (sec (x))^{4} d x

\frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + C

Solution steps

\int tan (x) (sec (x))^{4} d x

Simplify

= \int sec^{4} (x) tan (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u (1 + u^{2}) : u + u^{\frac{5}{2}}

= \int u + u^{\frac{5}{2}} 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 u d u + \int u^{\frac{5}{2}} d u

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

= \frac{2}{3} u^{\frac{3}{2}} + \frac{2}{7} u^{\frac{7}{2}}

Substitute back

u = tan (x)

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x)

Add a constant to the solution

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + C

\frac{d}{d x} (\frac{x ^{2}}{( x - 1 ) ^{2}})

\frac{d}{d x} (5 sec (x) tan (x))

u \to 8 lim (\frac{( u ^{2} - 5 u - 24 )}{u - 8})

\frac{\partial}{\partial x} (3 + 200 x^{2} + 300 y^{2})

f (x) = \frac{( 2 x - 1 )}{x + 3}, at x = 2

What is the integral of sqrt(tan(x))(sec(x))^4 ?
The integral of sqrt(tan(x))(sec(x))^4 is 2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x)+C