What is the integral of csc(4x+1)cot(4x+1) ?

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

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integral of csc(4x+1)cot(4x+1)

\int csc (4 x + 1) cot (4 x + 1) d x

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

Solution steps

\int csc (4 x + 1) cot (4 x + 1) d x

Apply u-substitution

: \int csc (u) cot (u) \frac{1}{4} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int - 1 d v

= \frac{1}{4} \cdot \int - 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{4} (- 1) v

Substitute back

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

Simplify

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

Add a constant to the solution

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

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