Cross Marsh
06/18/2023 · Senior High School

\( \int _{}^{}\frac{\sec (2x)^{2}}{\tan (2x)^{2}} d x \)

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La integral dada no se puede calcular directamente. Sin embargo, podemos simplificar la expresión antes de calcularla. Dado que \( \sec(2x) = \frac{1}{\cos(2x)} \) y \( \tan(2x) = \frac{\sin(2x)}{\cos(2x)} \), podemos reescribir la expresión como: \[ \frac{\sec (2x)^{2}}{\tan (2x)^{2}} = \frac{\left(\frac{1}{\cos(2x)}\right)^{2}}{\left(\frac{\sin(2x)}{\cos(2x)}\right)^{2}} = \frac{1}{\cos^{2}(2x)} \cdot \frac{\cos^{2}(2x)}{\sin^{2}(2x)} = \frac{1}{\sin^{2}(2x)} \] Por lo tanto, la integral se convierte en: \[ \int \frac{1}{\sin^{2}(2x)} dx \] Ahora podemos calcular esta integral. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int \frac{1}{\sin^{2}\left(2x\right)} dx\) - step1: Rewrite the expression: \(\int \csc^{2}\left(2x\right) dx\) - step2: Evaluate the integral: \(-\frac{1}{2}\cot\left(2x\right)\) - step3: Add the constant of integral C: \(-\frac{1}{2}\cot\left(2x\right) + C, C \in \mathbb{R}\) La integral de \( \frac{1}{\sin^{2}(2x)} \) es igual a \( -\frac{1}{2}\cot(2x) + C \), donde \( C \) es una constante arbitraria.

Quick Answer

La integral de \( \frac{\sec (2x)^{2}}{\tan (2x)^{2}} \) es \( -\frac{1}{2}\cot(2x) + C \).
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