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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u ∣ - \frac{u ^{2}}{2}

Substitute back

u = sin (x)

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

Add a constant to the solution

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

\int \frac{2 x + 4}{x ^{2} + 5 x - 6} d x

y = x^{3} - 7 x, y = - 4 x + 2

x \to 2 lim ((x + 4)^{2})

\frac{6 + sin ( x )}{6 x + cos ( x )}

\frac{\partial}{\partial y} (\frac{y}{2 x} - 1)

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