What is #f(x) = int tanx-secx dx# if #f(pi/4)=-1 #?

Question

What is #f(x) = int tanx-secx   dx# if #f(pi/4)=-1  #?

Evelyn Agard · Accepted Answer

Axel Alvarez · Answer

undefined To find $ f(x) = \int \tan(x) - \sec(x) \, dx $ if $ f(\frac{\pi}{4}) = -1 $, we need to first integrate the given function.

$ \int \tan(x) - \sec(x) \, dx = \int \tan(x) \, dx - \int \sec(x) \, dx $

The antiderivative of $ \tan(x) $ is $ -\ln|\cos(x)| $ and the antiderivative of $ \sec(x) $ is $ \ln|\sec(x) + \tan(x)| $.

So, $ f(x) = -\ln|\cos(x)| - \ln|\sec(x) + \tan(x)| + C $, where $ C $ is the constant of integration.

Given $ f(\frac{\pi}{4}) = -1 $, we substitute $ x = \frac{\pi}{4} $ into the function:

$ -1 = -\ln|\cos(\frac{\pi}{4})| - \ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| + C $

$ -1 = -\ln|\frac{1}{\sqrt{2}}| - \ln|\frac{1+\sqrt{2}}{\sqrt{2}}| + C $

$ -1 = -\ln\left(\frac{1}{\sqrt{2}}\right) - \ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right) + C $

$ -1 = -\ln(1) + \frac{1}{2}\ln(2) - \ln(1+\sqrt{2}) + C $

$ -1 = \frac{1}{2}\ln(2) - \ln(1+\sqrt{2}) + C $

Now, we solve for $ C $:

$ C = -1 - \frac{1}{2}\ln(2) + \ln(1+\sqrt{2}) $

Thus, the function $ f(x) = \int \tan(x) - \sec(x) \, dx $ with $ f(\frac{\pi}{4}) = -1 $ is:

$ f(x) = -\ln|\cos(x)| - \ln|\sec(x) + \tan(x)| - 1 - \frac{1}{2}\ln(2) + \ln(1+\sqrt{2}) $

Luke Alvarez · Answer

undefined To find $ f(x) = \int \tan(x) - \sec(x) \, dx $ when $ f\left(\frac{\pi}{4}\right) = -1 $, we first need to find the antiderivative of $ \tan(x) - \sec(x) $ and then evaluate it at $ x = \frac{\pi}{4} $.

The antiderivative of $ \tan(x) $ is $ -\ln|\cos(x)| $, and the antiderivative of $ \sec(x) $ is $ \ln|\sec(x) + \tan(x)| $.

So, the antiderivative of $ \tan(x) - \sec(x) $ is $ -\ln|\cos(x)| - \ln|\sec(x) + \tan(x)| $.

We need to evaluate this antiderivative at $ x = \frac{\pi}{4} $:

$ f\left(\frac{\pi}{4}\right) = -\ln|\cos\left(\frac{\pi}{4}\right)| - \ln|\sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right)| $

Now, $ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $, $ \sec\left(\frac{\pi}{4}\right) = \sqrt{2} $, and $ \tan\left(\frac{\pi}{4}\right) = 1 $.

So, $ f\left(\frac{\pi}{4}\right) = -\ln\left|\frac{1}{\sqrt{2}}\right| - \ln\left|\sqrt{2} + 1\right| $

Simplify:

$ f\left(\frac{\pi}{4}\right) = -\ln\left(\frac{1}{\sqrt{2}}\right) - \ln\left(\sqrt{2} + 1\right) $

Now, recall that $ \ln\left(\frac{1}{a}\right) = -\ln(a) $.

So, $ f\left(\frac{\pi}{4}\right) = \ln(\sqrt{2}) - \ln(\sqrt{2} + 1) $

$ f\left(\frac{\pi}{4}\right) = \ln\left(\frac{\sqrt{2}}{\sqrt{2} + 1}\right) $

Given that $ f\left(\frac{\pi}{4}\right) = -1 $, we have:

$ \ln\left(\frac{\sqrt{2}}{\sqrt{2} + 1}\right) = -1 $

Taking the exponential of both sides to solve for the fraction:

$ e^{-1} = \frac{\sqrt{2}}{\sqrt{2} + 1} $

Solve for the fraction:

$ e^{-1}(\sqrt{2} + 1) = \sqrt{2} $

$ e^{-1}\sqrt{2} + e^{-1} = \sqrt{2} $

$ e^{-1}\sqrt{2} = \sqrt{2} - e^{-1} $

Divide both sides by $ \sqrt{2} $:

$ e^{-1} = \frac{\sqrt{2} - e^{-1}}{\sqrt{2}} $

$ e^{-1} = \frac{\sqrt{2}}{\sqrt{2}} - \frac{e^{-1}}{\sqrt{2}} $

$ e^{-1} = 1 - \frac{e^{-1}}{\sqrt{2}} $

Add $ \frac{e^{-1}}{\sqrt{2}} $ to both sides:

$ e^{-1} + \frac{e^{-1}}{\sqrt{2}} = 1 $

Factor out $ e^{-1} $:

$ e^{-1}(1 + \frac{1}{\sqrt{2}}) = 1 $

Now, we solve for $ e^{-1} $:

$ e^{-1} = \frac{1}{1 + \frac{1}{\sqrt{2}}} $

Multiply the numerator and the denominator by $ \sqrt{2} $:

$ e^{-1} = \frac{\sqrt{2}}{\sqrt{2} + 1} $

Compare this result to the previous expression for $ e^{-1} $:

$ e^{-1} = \frac{\sqrt{2}}{\sqrt{2} + 1} $

Therefore, we've verified that $ e^{-1} = \frac{\sqrt{2}}{\sqrt{2} + 1} $.

Hence, $ f\left(\frac{\pi}{4}\right) = -1 $.