# What is #f(x) = int 2sin4x+tanx dx# if #f(pi/4)=1 #?

Evaluate the constant of integration

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To find ( f(x) = \int (2\sin(4x) + \tan(x)) , dx ) when ( f\left(\frac{\pi}{4}\right) = 1 ), first integrate the function with respect to ( x ):

[ f(x) = \int (2\sin(4x) + \tan(x)) , dx ]

[ = -\frac{1}{2}\cos(4x) + \ln|\sec(x)| + C ]

Then, use the given condition ( f\left(\frac{\pi}{4}\right) = 1 ) to find the constant ( C ):

[ 1 = -\frac{1}{2}\cos\left(\frac{\pi}{4}\right) + \ln|\sec\left(\frac{\pi}{4}\right)| + C ]

[ 1 = -\frac{1}{2}\left(\frac{\sqrt{2}}{2}\right) + \ln\left|\sec\left(\frac{\pi}{4}\right)\right| + C ]

[ 1 = -\frac{\sqrt{2}}{4} + \ln\left|\sqrt{2}\right| + C ]

[ 1 = -\frac{\sqrt{2}}{4} + \frac{1}{2}\ln(2) + C ]

[ C = 1 + \frac{\sqrt{2}}{4} - \frac{1}{2}\ln(2) ]

Now, substitute ( C ) back into the expression for ( f(x) ):

[ f(x) = -\frac{1}{2}\cos(4x) + \ln|\sec(x)| + 1 + \frac{\sqrt{2}}{4} - \frac{1}{2}\ln(2) ]

This is the expression for ( f(x) ).

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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.

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