# Given #int e^x(tanx + 1 )secx dx = e^xf(x)+C#.Write f(x)satisfying above.How can you solve it ?

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Please explain how can i get f(x)

#int e^x(tanx + 1 )secx dx = e^xf(x)+C#

Please explain how can i get f(x)

Given that

we get

Notice: In general,

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# f(x)=secx #

And if we compare with the integrand, we have:

And by observation, we note that:

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To find ( f(x) ), we'll start by expanding the given integral:

[ \int e^x(\tan(x) + 1)\sec(x) , dx = e^x f(x) + C ]

[ = \int e^x(\tan(x)\sec(x) + \sec(x)) , dx ]

[ = \int e^x(\sin(x) + 1) , dx ]

[ = \int e^x \sin(x) , dx + \int e^x , dx ]

[ = e^x \sin(x) - \int e^x \cos(x) , dx + e^x + C ]

[ = e^x (\sin(x) + 1) - \int e^x \cos(x) , dx + C ]

Now, let's denote ( f(x) ) as ( \sin(x) + 1 - \cos(x) ).

So, ( f(x) = \sin(x) + 1 - \cos(x) ).

This satisfies the given equation ( \int e^x(\tan(x) + 1)\sec(x) , dx = e^x f(x) + C ).

To solve it, we integrated by parts and used the trigonometric identity ( \sec(x) = \sin(x) + \cos(x) ) to simplify the integral. Then we solved the resulting integrals and rearranged the terms to find ( 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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