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Lelu [443]
3 years ago
7

Can anyone help me integrate :

Mathematics
1 answer:
worty [1.4K]3 years ago
3 0
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt

We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C

Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

\sec t=\dfrac{x+2}{\sqrt3}
\sec^2t=\dfrac{(x+2)^2}3
\tan t=\sqrt{\dfrac{x^2+4x+1}3}
\cot t=\sqrt{\dfrac3{x^2+4x+1}}
\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
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700+a=3944-b=c-80=3000​
velikii [3]

Answer:

A = 3244

B = 864

C = 3080

Step-by-step explanation:

 700 + a = 3944

 -700          -700

a = 3244

 3944 - b = c - 80 = 3000

c - 80 = 3000

  +80       +80

c = 3080

3944 - b = c ---> 3944 - b = 3080

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3 years ago
Tees and Cops
WINSTONCH [101]

If altogether they spent a total of $71,  and a cap cost $5, then the price of a t-shirt is $8. Option A is correct

From the question, we are told that all caps cost $5 each

Let the cost of each shirt be x:

If Andy bought a t-shirt and a cap for each of his six friends, there will be 6 shirts and 6 caps

If <u>Bob returned two caps and two t-shirts</u>, there will be 4 caps and 4 shirts left

If <u>Carl bought three t-shirts and returned a cap,</u> the number of caps remaining will be 3 caps and the number of shirts remaining will be 7 shirts

If they spent a total of $71, this means;

3(5) + 7x = 71

15 + 7x = 71

7x = 71 - 15

7x = 56

x = 56/7

x = 8

This shows that the price of a t-shirt is $8

Learn more here: brainly.com/question/16510024

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