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

The integral of (sec(x))^4 is 1/3 tan^3(x)+tan(x)+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))^4

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int v^{2} + 1 d v

= \int v^{2} + 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int v^{2} d v + \int 1 d v

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

\int 1 d v = v

= \frac{v ^{3}}{3} + v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int 16 + 9 x^{2} d x

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

f (x) = x + \frac{1}{x}

\int sin (x) (1 - cos^{2} (x)) d x

\int 5 cot^{3} (x) csc^{3} (x) d x

What is the integral of (sec(x))^4 ?
The integral of (sec(x))^4 is 1/3 tan^3(x)+tan(x)+C