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2 years ago

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In this instance, the statement, "Is the water cold?" is not a proposition. Therefore, <u>D. No</u>, because the statement does not make a claim; it is a question.

<h3>What is a proposition?</h3>

A proposition is a declarative statement that makes a claim that is either true or false.

Every proposition requires a solution or proof to find it true or not true.

Propositions need to be considered and debated before acceptance or rejection.

Thus, <u>Option D.</u> No, because the statement does not make a claim; it is a question.

Learn more about propositions at brainly.com/question/14766805

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Evaluate the integral following

alina1380 [7]

Answer:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

Step-by-step explanation:

We are given the integral of:

$\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}$

First, we can use a property to separate a constant out of integrand:

$\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}$

Next, expand the expression (integrand):

$\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}$

Since $\displaystyle{\sec x = \dfrac{1}{\cos x}}$ then it can be simplified to:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}$

Recall the formula:

$\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}$

For $\displaystyle{\cos ^2 x}$ , we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

$\displaystyle{2\cos^2 x -1 = \cos2x}$

We can solve for $\displaystyle{\cos ^2x}$ which is:

$\displaystyle{2\cos^2 x = \cos2x+1}\\\\\displaystyle{\cos^2x = \dfrac{\cos 2x +1}{2}}$

Therefore, we can write new integral as:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \dfrac{\cos2x +1}{2} \, dx}$

Evaluate each integral, applying the integration formula:

$\displaystyle{\int \dfrac{1}{\cos^2x} \, dx = \boxed{\tan x + C}}\\\\\displaystyle{\int -2 \, dx = \boxed{-2x + C}}\\\\\displaystyle{\int \dfrac{\cos 2x +1}{2} \, dx = \dfrac{1}{2}\int \cos 2x +1 \, dx}\\\\\displaystyle{= \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2x + x\right) + C}\\\\\displaystyle{= \boxed{\dfrac{1}{4}\sin 2x + \dfrac{1}{2}x + C}}$