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Exercise

$\frac{-2x\csc\left(2x\right)-2x^3\cos\left(2x\right)}{\sec\left(2x\right)}$

Step-by-step Solution

1

Expand the fraction $\frac{-2x\csc\left(2x\right)-2x^3\cos\left(2x\right)}{\sec\left(2x\right)}$ into $2$ simpler fractions with common denominator $\sec\left(2x\right)$

$\frac{-2x\csc\left(2x\right)}{\sec\left(2x\right)}+\frac{-2x^3\cos\left(2x\right)}{\sec\left(2x\right)}$
2

Applying the trigonometric identity: $\displaystyle\frac{1}{\sec(\theta)}=\cos(\theta)$

$-2x\cos\left(2x\right)\csc\left(2x\right)+\frac{-2x^3\cos\left(2x\right)}{\sec\left(2x\right)}$
3

Applying the trigonometric identity: $\displaystyle\frac{1}{\sec(\theta)}=\cos(\theta)$

$-2x\cos\left(2x\right)\csc\left(2x\right)-2x^3\cos\left(2x\right)\cos\left(2x\right)$
4

When multiplying two powers that have the same base ($\cos\left(2x\right)$), you can add the exponents

$-2x\cos\left(2x\right)\csc\left(2x\right)-2x^3\cos\left(2x\right)^2$
5

Simplify $-2x\cos\left(2x\right)\csc\left(2x\right)$ into $\cot(2x)$ by applying trigonometric identities

$-2x\cot\left(2x\right)-2x^3\cos\left(2x\right)^2$

Final answer to the exercise

$-2x\cot\left(2x\right)-2x^3\cos\left(2x\right)^2$

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