What is the integral of (cos^5(x))/(sin(x)) ?

The integral of (cos^5(x))/(sin(x)) is ln|sin(x)|-sin^2(x)+(sin^4(x))/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos^5(x))/(sin(x))

integral

\frac{cos ^{5} ( x )}{sin ( x )}

ln ∣ sin (x) ∣ - sin^{2} (x) + \frac{sin ^{4} ( x )}{4} + C

Solution steps

\int \frac{cos ^{5} ( x )}{sin ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int 2 u d u = u^{2}

\int u^{3} d u = \frac{u ^{4}}{4}

= ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4}

Substitute back

u = sin (x)

= ln ∣ sin (x) ∣ - sin^{2} (x) + \frac{sin ^{4} ( x )}{4}

Add a constant to the solution

= ln ∣ sin (x) ∣ - sin^{2} (x) + \frac{sin ^{4} ( x )}{4} + C

y^{\frac{- 6}{7}}

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

x \cdot (x + 5)^{8}

tan^{3} (2 x + 1)

(sin (x) + 3)^{- 1}

What is the integral of (cos^5(x))/(sin(x)) ?
The integral of (cos^5(x))/(sin(x)) is ln|sin(x)|-sin^2(x)+(sin^4(x))/4+C