What is the integral of 8csc^2(4x)cot(4x) ?

The integral of 8csc^2(4x)cot(4x) is -cot^2(4x)+C

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integral of 8csc^2(4x)cot(4x)

\int 8 csc^{2} (4 x) cot (4 x) d x

- cot^{2} (4 x) + C

Solution steps

\int 8 csc^{2} (4 x) cot (4 x) d x

Take the constant out:

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

= 8 \cdot \int csc^{2} (4 x) cot (4 x) d x

Apply u-substitution

: \int - \frac{u}{4} d u

= 8 \cdot \int - \frac{u}{4} d u

Take the constant out:

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

= 8 (- \frac{1}{4} \cdot \int u d u)

Apply the Power Rule

: \frac{u ^{2}}{2}

= 8 (- \frac{1}{4} \cdot \frac{u ^{2}}{2})

Substitute back

u = cot (4 x)

= 8 (- \frac{1}{4} \cdot \frac{cot ^{2} ( 4 x )}{2})

Simplify

8 (- \frac{1}{4} \cdot \frac{cot ^{2} ( 4 x )}{2}) : - cot^{2} (4 x)

= - cot^{2} (4 x)

Add a constant to the solution

= - cot^{2} (4 x) + C

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What is the integral of 8csc^2(4x)cot(4x) ?
The integral of 8csc^2(4x)cot(4x) is -cot^2(4x)+C