What is the integral of (sec^4(x))/(tan^3(x)) ?

The integral of (sec^4(x))/(tan^3(x)) is -1/2 cot^2(x)+ln|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^4(x))/(tan^3(x))

\int \frac{sec ^{4} ( x )}{tan ^{3} ( x )} d x

- \frac{1}{2} cot^{2} (x) + ln ∣ tan (x) ∣ + C

Solution steps

\int \frac{sec ^{4} ( x )}{tan ^{3} ( x )} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{sec ^{4} ( x )}{tan ^{3} ( x )} = (sec^{4} (x)) tan^{- 3} (x)

= \int sec^{4} (x) tan^{- 3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{1}{2 u ^{2}} + ln ∣ u ∣

Substitute back

u = tan (x)

= - \frac{1}{2 tan ^{2} ( x )} + ln ∣ tan (x) ∣

Simplify

- \frac{1}{2 tan ^{2} ( x )} + ln ∣ tan (x) ∣ : - \frac{1}{2} cot^{2} (x) + ln ∣ tan (x) ∣

= - \frac{1}{2} cot^{2} (x) + ln ∣ tan (x) ∣

Add a constant to the solution

= - \frac{1}{2} cot^{2} (x) + ln ∣ tan (x) ∣ + C

\int (x - 5) cos (x) d x

x \to 2 lim (\frac{( x - 2 )}{x - 2})

\int \frac{3}{x ^{2} - 1} d x

\int θ sec^{2} (θ) d θ

sin (x^{4})

What is the integral of (sec^4(x))/(tan^3(x)) ?
The integral of (sec^4(x))/(tan^3(x)) is -1/2 cot^2(x)+ln|tan(x)|+C