Cross Marsh
06/18/2023 · Senior High School
\( \int _{}^{}\frac{\sec (2x)^{2}}{\tan (2x)^{2}} d x \)
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Step-by-step Solution
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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