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

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

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

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

- cot (x) - x + C

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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

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

\int 1 d x = x

= - cot (x) - x

Add a constant to the solution

= - cot (x) - x + C

\int \frac{1}{x ^{2} + 4 x + 10} d x

y cos^{2} (x) \frac{d y}{d x} = 4 + tan^{2} (x)

\int x + 7 d x

2 x^{3} - 3 x^{2} - 72 x + 108

ln (\frac{( 4 - x )}{4 + x})

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