How do you integrate #int4sec4x*tan4x*sec^4 4x# using substitution?

Answer 1

Well, let's rewrite this, and you may see something.

Let #u = 4x#, and you have #du = 4dx#.
#color(blue)(int 4sec4xtan4xsec^4 (4x)dx)#
#= int secutanusec^4(u)du#
So now... let #v = secu#, and you get #dv = secutanudu#:
#=> int (secu)^4secutanudu#
#= int v^4dv#
#= v^5/5#

Back-substitute to get:

#= sec^5(u)/5#
#= color(blue)(sec^5(4x)/5 + C)#
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Answer 2

To integrate ∫4sec^4(x)tan(x)sec^4(4x) dx using substitution, we can use the substitution method.

Let's set u = sec(x), then du = sec(x)tan(x) dx.

Now, we need to express everything in terms of u:

sec^2(x) = u^2 - 1

tan(x) = √(u^2 - 1)

sec(4x) = 1/cos(4x) = 1/(1 - tan^2(4x)) = 1/(1 - (u^2 - 1)^2)

Therefore, our integral becomes:

∫4sec^4(x)tan(x)sec^4(4x) dx = ∫4u^4√(u^2 - 1) * 1/(1 - (u^2 - 1)^2) du

Now, we need to simplify this expression and then integrate with respect to u. This process may involve trigonometric identities and algebraic manipulations.

After integrating with respect to u, we'll have the antiderivative in terms of u. Finally, we'll substitute back u = sec(x) to express the final result in terms of 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.

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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