How do you find the antiderivative of #(cosx+secx)^2#?

Answer 1

# int \ (cosx+secx)^2 \ dx = sin(2x)/4 + 5/2x + tanx + C #

We seek:

# I = int \ (cosx+secx)^2 \ dx #

Which we can write:

# I = int \ cos^2x + 2cosxsecx + sec^2x \ dx # # \ \ = int \ cos^2x + 2 + sec^2x \ dx # # \ \ = int \ (cos(2x)+1)/2 + 2 + sec^2x \ dx # # \ \ = int \ cos(2x)/2 + 5/2 + sec^2x \ dx #

All integrand functions are standard results, so we can now readily integrate to get:

# I = sin(2x)/4 + 5/2x + tanx + C #
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Answer 2

To find the antiderivative of ((\cos(x) + \sec(x))^2), you can use the trigonometric identity (\sec^2(x) = \tan^2(x) + 1), then perform a substitution. Let (u = \cos(x) + \sec(x)). Then, (du = (-\sin(x) + \sec(x)\tan(x))dx). Rearranging terms, (-du = (\sec(x)\tan(x) - \sin(x))dx). Now, rewrite the integral in terms of (u): (\int -\frac{1}{\sec(x)\tan(x) - \sin(x)} du). This integral can be evaluated using partial fraction decomposition. After integrating, substitute (u) back in to get the antiderivative 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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