Howell Brewer
03/13/2024 · High School
3. (2,0) Calcular o comprimento de arco da curva \( y=1-\ln (\operatorname{sen} x) \) do ponto onde \( x=\frac{\pi}{6} \) ao ponto onde \( x=\frac{\pi}{4} \)
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Desculpe, mas parece que não é possível calcular essa integral diretamente. Vamos tentar simplificar a expressão antes de calcular a integral.
Vamos simplificar a expressão \( \sqrt{1 + \left( -\frac{1}{\operatorname{sen} x} \cdot \operatorname{cos} x \right)^2} \) antes de calcular a integral.
Simplify the expression by following steps:
- step0: Solution:
\(\sqrt{1+\left(-\frac{1}{\sin\left(x\right)}\times \cos\left(x\right)\right)^{2}}\)
- step1: Multiply the terms:
\(\sqrt{1+\left(-\frac{\cos\left(x\right)}{\sin\left(x\right)}\right)^{2}}\)
- step2: Rewrite the expression:
\(\sqrt{\frac{\sin^{2}\left(x\right)+\cos^{2}\left(x\right)}{\sin^{2}\left(x\right)}}\)
- step3: Use the properties of radicals:
\(\frac{\sqrt{\sin^{2}\left(x\right)+\cos^{2}\left(x\right)}}{\sqrt{\sin^{2}\left(x\right)}}\)
- step4: Simplify the expression:
\(\frac{1}{\sin\left(x\right)}\)
- step5: Transform the expression:
\(\csc\left(x\right)\)
Vamos substituir a expressão simplificada na integral:
\[ L = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc(x) \, dx \]
Agora, vamos calcular essa integral para encontrar o comprimento de arco da curva.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
\(\int_{\frac{\pi }{6}}^{\frac{\pi }{4}} \csc\left(x\right) dx\)
- step1: Evaluate the integral:
\(\int \csc\left(x\right) dx\)
- step2: Evaluate the integral:
\(-\ln{\left(\left|\csc\left(x\right)+\cot\left(x\right)\right|\right)}\)
- step3: Return the limits:
\(\left(-\ln{\left(\left|\csc\left(x\right)+\cot\left(x\right)\right|\right)}\right)\bigg |_{\frac{\pi }{6}}^{\frac{\pi }{4}}\)
- step4: Calculate the value:
\(\ln{\left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right)}\)
O comprimento de arco da curva \( y=1-\ln (\operatorname{sen} x) \) do ponto onde \( x=\frac{\pi}{6} \) ao ponto onde \( x=\frac{\pi}{4} \) é \( \ln{(\\frac{2+\\sqrt{3}}{\\sqrt{2}+1})} \).
Quick Answer
O comprimento de arco da curva \( y=1-\ln (\operatorname{sen} x) \) entre \( x=\frac{\pi}{6} \) e \( x=\frac{\pi}{4} \) é \( \ln\left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right) \).
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