How do you integrate #intsec3x#?

Answer 1

#intsec(3x)dx=1/3lnabs(sec(3x)+tan(3x))+C#

#intsec(3x)dx#
Use the substitution #u=3x#, implying that #du=(3)dx#. Then:
#=1/3intsec(3x)(3)dx=1/3intsec(u)du#
This is a common integral: #intsec(u)du=lnabs(sec(u)+tan(u))+C#
We can derive this integral by multiplying the integrand by #(sec(u)+tan(u))/(sec(u)+tan(u))#:
#=1/3intsec(u)(sec(u)+tan(u))/(sec(u)+tan(u))du=1/3int(sec^2(u)+sec(u)tan(u))/(sec(u)+tan(u))du#
Now let #v=sec(u)+tan(u)#. This implies that #dv=(sec(u)tan(u)+sec^2(u))du#. This is the numerator:
#=1/3int(dv)/v=1/3lnabsv+C=1/3lnabs(sec(u)+tan(u))+C#

Finally:

#=1/3lnabs(sec(3x)+tan(3x))+C#
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Answer 2

To integrate the function (\int \sec^3(x) , dx), we can use trigonometric substitution. We can rewrite (\sec^3(x)) as (\sec(x) \cdot \sec^2(x)). Then, using the identity (\sec^2(x) = \tan^2(x) + 1), we can express (\sec^3(x)) in terms of (\tan(x)). After substitution, we can integrate using the properties of trigonometric functions.

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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.

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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