What is the integral of (cos^5(x))/(sin^4(x)) ?

The integral of (cos^5(x))/(sin^4(x)) is -1/3 csc^3(x)+2csc(x)+sin(x)+C

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integral of (cos^5(x))/(sin^4(x))

\int \frac{cos ^{5} ( x )}{sin ^{4} ( x )} d x

- \frac{1}{3} csc^{3} (x) + 2 csc (x) + sin (x) + C

Solution steps

\int \frac{cos ^{5} ( x )}{sin ^{4} ( x )} d x

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( x ) ) ^{2} cos ( x )}{sin ^{4} ( x )} d x

Apply u-substitution

: \int \frac{( 1 - u ^{2} ) ^{2}}{u ^{4}} d u

= \int \frac{( 1 - u ^{2} ) ^{2}}{u ^{4}} d u

Expand

\frac{( 1 - u ^{2} ) ^{2}}{u ^{4}} : \frac{1}{u ^{4}} - \frac{2}{u ^{2}} + 1

= \int \frac{1}{u ^{4}} - \frac{2}{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 ^{4}} d u - \int \frac{2}{u ^{2}} d u + \int 1 d u

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

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

\int 1 d u = u

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

Substitute back

u = sin (x)

= - \frac{1}{3 sin ^{3} ( x )} - (- \frac{2}{sin ( x )}) + sin (x)

Simplify

- \frac{1}{3 sin ^{3} ( x )} - (- \frac{2}{sin ( x )}) + sin (x) : - \frac{1}{3} csc^{3} (x) + 2 csc (x) + sin (x)

= - \frac{1}{3} csc^{3} (x) + 2 csc (x) + sin (x)

Add a constant to the solution

= - \frac{1}{3} csc^{3} (x) + 2 csc (x) + sin (x) + C

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What is the integral of (cos^5(x))/(sin^4(x)) ?
The integral of (cos^5(x))/(sin^4(x)) is -1/3 csc^3(x)+2csc(x)+sin(x)+C