What is #f(x) = int tanx-secx dx# if #f(pi/4)=-1 #?
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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}) )
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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 ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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