What is the integral of (cos^5(x))/(sin^3(x)) ?

The integral of (cos^5(x))/(sin^3(x)) is -1/2 csc^2(x)-2ln|sin(x)|+1/2 sin^2(x)+C

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integral of (cos^5(x))/(sin^3(x))

\int \frac{cos ^{5} ( x )}{sin ^{3} ( x )} d x

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

Solution steps

\int \frac{cos ^{5} ( x )}{sin ^{3} ( x )} d x

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( x ) ) ^{2} cos ( x )}{sin ^{3} ( x )} d x

Apply u-substitution

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

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

Expand

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

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

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

Substitute back

u = sin (x)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (cos^5(x))/(sin^3(x)) ?
The integral of (cos^5(x))/(sin^3(x)) is -1/2 csc^2(x)-2ln|sin(x)|+1/2 sin^2(x)+C