# What is #f(x) = int sin2x-5secx dx# if #f(pi/4)=-2 #?

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Given that (f(\frac{\pi}{4}) = -2), we need to find (f(x)) for the function (f(x) = \int (\sin(2x) - 5\sec(x)) , dx).

We know that (f'(x) = \sin(2x) - 5\sec(x)).

Using the given initial condition (f(\frac{\pi}{4}) = -2), we can solve for (f(x)) by integrating (f'(x)) with respect to (x) and applying the initial condition.

So, integrating (f'(x)) gives:

[ f(x) = -\frac{1}{2}\cos(2x) - 5\ln|\sec(x) + \tan(x)| + C ]

Using the initial condition (f(\frac{\pi}{4}) = -2), we can find (C):

[ -2 = -\frac{1}{2}\cos\left(\frac{\pi}{2}\right) - 5\ln\left|\sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right)\right| + C ]

Solve for (C):

[ C = -2 + \frac{1}{2} - 5\ln\left|1 + 1\right| = -\frac{3}{2} ]

Therefore, the function (f(x)) is:

[ f(x) = -\frac{1}{2}\cos(2x) - 5\ln|\sec(x) + \tan(x)| - \frac{3}{2} ]

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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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