What is the integral of 2xcsc(2x)cot(2x) ?

The integral of 2xcsc(2x)cot(2x) is 1/2 (-2xcsc(2x)+ln|tan(x)|)+C

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integral of 2xcsc(2x)cot(2x)

\int 2 x csc (2 x) cot (2 x) d x

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

Solution steps

\int 2 x csc (2 x) cot (2 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x csc (2 x) cot (2 x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{u cot ( u )}{4 sin ( u )} d u

= 2 \cdot \int \frac{u cot ( u )}{4 sin ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{4} \cdot \int \frac{u cot ( u )}{sin ( u )} d u

Rewrite using trig identities

= 2 \cdot \frac{1}{4} \cdot \int u cot (u) csc (u) d u

Apply Integration By Parts

: - u csc (u) - \int - csc (u) d u

= 2 \cdot \frac{1}{4} (- u csc (u) - \int - csc (u) d u)

\int - csc (u) d u = - ln tan (\frac{u}{2})

= 2 \cdot \frac{1}{4} (- u csc (u) - (- ln tan (\frac{u}{2})))

Substitute back

u = 2 x

= 2 \cdot \frac{1}{4} (- 2 x csc (2 x) - (- ln tan (\frac{2 x}{2})))

Simplify

2 \cdot \frac{1}{4} (- 2 x csc (2 x) - (- ln tan (\frac{2 x}{2}))) : \frac{1}{2} (- 2 x csc (2 x) + ln ∣ tan (x) ∣)

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

Add a constant to the solution

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

\int x^{x} d x

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

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

\frac{\partial}{\partial x} (5 x^{4} y^{2})

\frac{\partial}{\partial y} (\frac{5 x}{y})

What is the integral of 2xcsc(2x)cot(2x) ?
The integral of 2xcsc(2x)cot(2x) is 1/2 (-2xcsc(2x)+ln|tan(x)|)+C