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^2(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^{2} (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{u}{u ^{2} + 1}

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u ∣ - \frac{1}{2} ln u^{2} + 1

Substitute back

u = sin (x)

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

Add a constant to the solution

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

\int \frac{x ^{2} - 1}{2 x ^{2} - 32} d x

\int \frac{x ^{2} + 7}{x ^{2} + 9} d x

\int (2 x^{2} - x) cos (4 x) d x

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

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

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^2(x)+1|+C