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kramer
3 years ago
7

Find the area of a regular hexagon

Mathematics
2 answers:
8090 [49]3 years ago
6 0

Answer:

25 cm

Step-by-step explanation:

andreyandreev [35.5K]3 years ago
6 0

Answer: 25 would be the answer

Step-by-step explanation: Hope this helps vote me brainliest

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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}}

Then add all these boxed integrated together then we'll get:

\displaystyle{4\left(\tan x - 2x + \dfrac{1}{4}\sin 2x + \dfrac{1}{2} x\right) + C}

Expand 4 in the expression:

\displaystyle{4\tan x - 8x +\sin 2x + 2 x + C}\\\\\displaystyle{4\tan x + \sin 2x - 6x + C}

Therefore, the answer is:

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

4 0
1 year ago
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