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

The integral of (cos^2(2x))/(sin(2x)) is 1/2 ln|tan(x)|+1/2 cos(2x)+C

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

integral

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

\frac{1}{2} ln ∣ tan (x) ∣ + \frac{1}{2} cos (2 x) + C

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

\int \frac{1}{sin ( 2 x )} d x = \frac{1}{2} ln ∣ tan (x) ∣

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

= \frac{1}{2} ln ∣ tan (x) ∣ - (- \frac{1}{2} cos (2 x))

Simplify

= \frac{1}{2} ln ∣ tan (x) ∣ + \frac{1}{2} cos (2 x)

Add a constant to the solution

= \frac{1}{2} ln ∣ tan (x) ∣ + \frac{1}{2} cos (2 x) + C

- 3 sin (x)

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

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

(e^{x})^{2} x^{2}

\frac{x ^{2}}{4 - x ^{2}}

What is the integral of (cos^2(2x))/(sin(2x)) ?
The integral of (cos^2(2x))/(sin(2x)) is 1/2 ln|tan(x)|+1/2 cos(2x)+C