What is the integral of (cot(2x+1))/(sin^2(2x+1)) ?

The integral of (cot(2x+1))/(sin^2(2x+1)) is -1/4 csc^2(2x+1)+C

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integral of (cot(2x+1))/(sin^2(2x+1))

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

- \frac{1}{4} csc^{2} (2 x + 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{cot ( u )}{2 sin ^{2} ( u )} d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1}{2} d v

= \frac{1}{2} \cdot \int - \frac{1}{2} d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} (- \frac{1}{2}) v

Substitute back

= \frac{1}{2} (- \frac{1}{2}) csc^{2} (2 x + 1)

Simplify

\frac{1}{2} (- \frac{1}{2}) csc^{2} (2 x + 1) : - \frac{1}{4} csc^{2} (2 x + 1)

= - \frac{1}{4} csc^{2} (2 x + 1)

Add a constant to the solution

= - \frac{1}{4} csc^{2} (2 x + 1) + C

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What is the integral of (cot(2x+1))/(sin^2(2x+1)) ?
The integral of (cot(2x+1))/(sin^2(2x+1)) is -1/4 csc^2(2x+1)+C