What is the integral of (5sec(θ)tan(θ))/(sec^2(θ)-sec(θ)) ?

The integral of (5sec(θ)tan(θ))/(sec^2(θ)-sec(θ)) is 5(-ln|sec(θ)|+ln|sec(θ)-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5sec(θ)tan(θ))/(sec^2(θ)-sec(θ))

\int \frac{5 sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

5 (- ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣) + C

Solution steps

\int \frac{5 sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int \frac{sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

Apply u-substitution

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

= 5 \cdot \int \frac{1}{u ^{2} - u} d u

Take the partial fraction of

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

= 5 \cdot \int - \frac{1}{u} + \frac{1}{u - 1} 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

= 5 (- \int \frac{1}{u} d u + \int \frac{1}{u - 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= 5 (- ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = sec (θ)

= 5 (- ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣)

Add a constant to the solution

= 5 (- ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣) + C

\int \frac{3 ^{t a n (5 x)}}{cos ^{2} ( 5 x )} d x

\frac{( t ^{3} + t )}{t ^{4} - 2}

x \to 2 + lim (f (x))

\int \frac{21}{4 + 25 x ^{2}} d x

n \to \infty lim (e^{n})

What is the integral of (5sec(θ)tan(θ))/(sec^2(θ)-sec(θ)) ?
The integral of (5sec(θ)tan(θ))/(sec^2(θ)-sec(θ)) is 5(-ln|sec(θ)|+ln|sec(θ)-1|)+C