What is the integral of (x+3)csc^2(x^2+6x-4) ?

The integral of (x+3)csc^2(x^2+6x-4) is -1/2 cot(x^2+6x-4)+C

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integral of (x+3)csc^2(x^2+6x-4)

\int (x + 3) csc^{2} (x^{2} + 6 x - 4) d x

- \frac{1}{2} cot (x^{2} + 6 x - 4) + C

Solution steps

\int (x + 3) csc^{2} (x^{2} + 6 x - 4) d x

Apply u-substitution

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

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

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{2} (- cot (u))

Substitute back

u = x^{2} + 6 x - 4

= \frac{1}{2} (- cot (x^{2} + 6 x - 4))

Simplify

= - \frac{1}{2} cot (x^{2} + 6 x - 4)

Add a constant to the solution

= - \frac{1}{2} cot (x^{2} + 6 x - 4) + C

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What is the integral of (x+3)csc^2(x^2+6x-4) ?
The integral of (x+3)csc^2(x^2+6x-4) is -1/2 cot(x^2+6x-4)+C