What is the integral of ((cos^3(x)))/(sin^4(x)) ?

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

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integral of ((cos^3(x)))/(sin^4(x))

integral

\frac{( cos ^{3} ( x ) )}{sin ^{4} ( x )}

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

Solution steps

\int \frac{cos ^{3} ( x )}{sin ^{4} ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{1 - u ^{2}}{u ^{4}} : \frac{1}{u ^{4}} - \frac{1}{u ^{2}}

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

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

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

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

Substitute back

u = sin (x)

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

Simplify

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

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

Add a constant to the solution

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

(t^{2})^{2}

\frac{1}{( 2 x - x ^{2} )}

\frac{3 x ^{2} + 2 x}{x ^{6}}

1 + 2 x

(1 + 2 cos (x))^{2}

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