What is the integral of tan^3(x)(sec(x))^{-1/2} ?

The integral of tan^3(x)(sec(x))^{-1/2} is 2/(sec^{1/2)(x)}+2/3 sec^{3/2}(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^3(x)(sec(x))^{-1/2}

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

\frac{2}{sec ^{\frac{1}{2}} ( x )} + \frac{2}{3} sec^{\frac{3}{2}} (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = sec (x)

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

Simplify

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

Add a constant to the solution

= \frac{2}{sec ^{\frac{1}{2}} ( x )} + \frac{2}{3} sec^{\frac{3}{2}} (x) + C

\int_{0}^{2} (2 + x) d x

5 e^{x} + x

\int k \cdot x d x

(x, y) \to (0, 4) lim (x^{3} y)

f (x) = ln (2), y = 0, y = 2

What is the integral of tan^3(x)(sec(x))^{-1/2} ?
The integral of tan^3(x)(sec(x))^{-1/2} is 2/(sec^{1/2)(x)}+2/3 sec^{3/2}(x)+C