What is the integral of (cos(x))/(sin^3(x)-sin(x)) ?

The integral of (cos(x))/(sin^3(x)-sin(x)) is -ln|sin(x)|+1/2 ln|sin(x)+1|+1/2 ln|sin(x)-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos(x))/(sin^3(x)-sin(x))

\int \frac{cos ( x )}{sin ^{3} ( x ) - sin ( x )} d x

- ln ∣ sin (x) ∣ + \frac{1}{2} ln ∣ sin (x) + 1 ∣ + \frac{1}{2} ln ∣ sin (x) - 1 ∣ + C

Solution steps

\int \frac{cos ( x )}{sin ^{3} ( x ) - sin ( x )} d x

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

= - ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

Substitute back

u = sin (x)

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

Add a constant to the solution

= - ln ∣ sin (x) ∣ + \frac{1}{2} ln ∣ sin (x) + 1 ∣ + \frac{1}{2} ln ∣ sin (x) - 1 ∣ + C

\int (sin (x) + 2) d x

\int 6 x (3 x^{2} + 5)^{6} d x

\int \frac{e ^{- t}}{t ^{2}} d t

\int e^{c o s (18 t)} sin (18 t) d t

\int cot^{2} (u) d u

What is the integral of (cos(x))/(sin^3(x)-sin(x)) ?
The integral of (cos(x))/(sin^3(x)-sin(x)) is -ln|sin(x)|+1/2 ln|sin(x)+1|+1/2 ln|sin(x)-1|+C