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

Please explain how can i get f(x)
#int e^x(tanx + 1 )secx dx = e^xf(x)+C#

Answer 1

#f(x)=\sec x#

Given that

#\int e^x(\tan x+1)\sec x\ dx#
#=\int e^x\tan x\sec x\ dx+\int e^x\sec x\ dx#
#=e^x\int \sec x\tan x\ dx-\int (\frac{d}{dx}e^x\cdot \int \sec x\tan x\ dx)dx+\int e^x\sec x\ dx#
#=e^x\sec x-\int (e^x\sec x)dx+\int e^x\sec x\ dx#
#=e^x\sec x+C#
#=e^xf(x)+C#

we get

#f(x)=\sec x#

Notice: In general,

#\int e^x(f(x)+f'(x))\ dx#
#=\int e^xf(x)dx+\int e^x f'(x)\ dx#
#=\int e^xf'(x)dx+\int e^x f(x)\ dx#
#=e^x\int f'(x)dx-\int(\frac{d}{dx} e^x\cdot \int f'(x)\ dx)dx+\int e^x f(x)\ dx#
#=e^xf(x)-\int e^xf(x)dx+\int e^x f(x)\ dx#
#=e^xf(x)+C#
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Answer 2

# f(x)=secx #

As we are given a suggested solution then the simpler approach is to differentiate that solution and compare. So we use the product rule to differentiate #e^xf(x)+C# to get
# d/dx (e^xf(x)+C) = e^x(d/dxf(x)) + (d/dxe^x)f(x) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = e^xf'(x) + e^xf(x) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = e^x(f'(x) + f(x)) #

And if we compare with the integrand, we have:

# e^x(f'(x) + f(x)) = e^x(tanx+1)secx#
# :. f'(x) + f(x) = (tanx+1)secx#
# :. f'(x) + f(x) = secxtanx+secx#

And by observation, we note that:

# d/dxsecx=secxtanx => f(x)=secx #
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Answer 3

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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Answer from HIX Tutor

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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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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