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Use property 8 to estimate the value of the integral.y tanxdx

User Serhio
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The estimated value of the integral y = ∫tanx dx is y = -ln(cos(x)) + c

How to estimate the value of the integral.

From the question, we have the following parameters that can be used in our computation:

y = ∫tanx dx

This means that


y = \int\limits {\tan(x)} \, dx

Rewrite as


y = \int\limits {(\sin(x))/(\cos(x)) \, dx

Let u = cos(x)

So: du = -sin(x) dx

So, we have


y = \int\limits {(-1)/(u) \, du

Factor out the -1


y = -\int\limits {(1)/(u) \, du

Integrate

y = -ln(u) + c

Recall that

u = cos(x)

So, we have

y = -ln(cos(x)) + c

Hence, the estimated value of the integral is y = -ln(cos(x)) + c

User AniV
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6 votes

The integral of
\(y \tan(x) \, dx\) is \(-y \ln|\cos(x)| + C\), where C is the constant of integration.

The integral
\(\int y \tan(x) \, dx\) can be solved using integration by substitution.

Given the integral
\(\int y \tan(x) \, dx\), let's use the substitution method where
\(u = \cos(x)\) and \(du = -\sin(x) \, dx\).

Rewriting the integral in terms of u:


\[ \int y \tan(x) \, dx = \int y (\sin(x))/(\cos(x)) \, dx = -y \int (1)/(u) \, du\]

This simplifies to:


\[ -y \int (1)/(u) \, du = -y \ln|u| + C = -y \ln|\cos(x)| + C\]

Therefore, the solution to the integral
\(\int y \tan(x) \, dx\) is \(-y \ln|\cos(x)| + C\), where C is the constant of integration.

User Lucas Van Dijk
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7.1k points
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