What is the integral of #(secxtanx)(1+secx)dx#?

Answer 1
#1/2 (1+secx)^2+C#
#int (secxtanx)(1+secx)dx=int (1+secx)(secxtanx)dx #
Let #u=1+secx#. Then #du=secxtanxdx#
#int udu=1/2u^2+C# So,
#int (secxtanx)(1+secx)dx=1/2(1+secx)^2+C#
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Answer 2

To find the integral of ((\sec(x)\tan(x))(1+\sec(x)),dx), first expand the expression inside the integral:

[ \int (\sec(x)\tan(x))(1+\sec(x)),dx = \int (\sec(x)\tan(x) + \sec^2(x)\tan(x)),dx ]

Now, integrate each term separately:

  1. For (\int \sec(x)\tan(x),dx), we know the integral of (\sec(x)\tan(x)) is (\sec(x)). This is a standard result from integration formulas.

  2. For (\int \sec^2(x)\tan(x),dx), let's use a substitution method. Let (u = \sec(x)), which implies (du = \sec(x)\tan(x),dx). The integral becomes:

[ \int u,du = \frac{1}{2}u^2 = \frac{1}{2}\sec^2(x) ]

So, combining both parts, the integral is:

[ \sec(x) + \frac{1}{2}\sec^2(x) + C ]

where (C) is the constant of integration.

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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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