What is the integral of (cos^3(x))/(sin^2(x)) ?

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

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integral of (cos^3(x))/(sin^2(x))

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

- csc (x) - sin (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int 1 d u = u

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

Substitute back

u = sin (x)

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

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= - csc (x) - sin (x)

Add a constant to the solution

= - csc (x) - sin (x) + C

\int (\frac{7}{x} + \frac{15}{x ^{6}}) d x

y = sin (2 x) cos (2 x)

\frac{\partial}{\partial x} ((\frac{x + 2}{y + 1})^{\frac{1}{2}})

e^{a s}

n \to \infty lim (- 3 + 4 n^{5})

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