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

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

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

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

- cot (x) - 2 x + C

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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

\int 2 d x = 2 x

= - cot (x) - 2 x

Add a constant to the solution

= - cot (x) - 2 x + C

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

\int \frac{1}{( x ^{2} + 2 x + 5 ) ^{2}} d x

\int \frac{1}{3 + 2 cos ( x )} d x

\int sin (2 x) cos (2 x) d x

\int e^{2 x + 1} d x

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