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Ber [7]
2 years ago
11

Evaluate the following integral using trigonometric substitution

Mathematics
1 answer:
serg [7]2 years ago
8 0

Answer:

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

Step-by-step explanation:

We are given the following integral:

\int \frac{dx}{\sqrt{9-x^2}}

Trigonometric substitution:

We have the term in the following format: a^2 - x^2, in which a = 3.

In this case, the substitution is given by:

x = a\sin{\theta}

So

dx = a\cos{\theta}d\theta

In this question:

a = 3

x = 3\sin{\theta}

dx = 3\cos{\theta}d\theta

So

\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}

We have the following trigonometric identity:

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

So

1 - \sin^{2}{\theta} = \cos^{2}{\theta}

Replacing into the integral:

\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C

Coming back to x:

We have that:

x = 3\sin{\theta}

So

\sin{\theta} = \frac{x}{3}

Applying the arcsine(inverse sine) function to both sides, we get that:

\theta = \arcsin{(\frac{x}{3})}

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

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Help!! Find missing value for each quadratic function
Svetllana [295]

I'll do the first one to get you started. The answer is -3

=================================================

Explanation:

When x = -1, y = 0 as shown in the first column. The second column says that (x,y) = (0,1) and the third column says (x,y) = (1,0)

We'll use these three points to determine the quadratic function that goes through them all.

The general template we'll use is y = ax^2 + bx + c

------------

Plug in (x,y) = (0,1). Simplify

y = ax^2 + bx + c

1 = a*0^2 + b*0 + c

1 = 0a + 0b + c

1 = c

c = 1

------------

Plug in (x,y) = (-1,0) and c = 1

y = ax^2 + bx + c

0 = a*(-1)^2 + b(-1) + 1

0 = 1a - 1b + 1

a-b+1 = 0

a-b = -1

a = b-1

------------

Plug in (x,y) = (1,0), c = 1, and a = b-1

y = ax^2 + bx + c

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0 = a + b + 1

0 = b-1 + b + 1 ... replace 'a' with b-1

0 = 2b

2b = 0

b = 0/2

b = 0

------------

If b = 0, then 'a' is...

a = b-1

a = 0-1

a = -1

------------

In summary so far, we found: a = -1, b = 0, c = 1

Therefore, y = ax^2 + bx + c turns into y = -1x^2 + 0x + 1 which simplifies to y = -x^2 + 1

------------

The last thing to do is plug x = 2 into this equation and simplify

y = -x^2 + 1

y = -2^2 + 1

y = -4 + 1

y = -3


3 0
3 years ago
???????anyone know...
RUDIKE [14]
The answer is:  " 60° " .
__________________________________________________________
     " m∠A = 60° " .
__________________________________________________________
Explanation:
__________________________________________________________

Note:  All triangles, by definition, have 3 (three) sides and 3 (three angles).

The triangle shown (in the "image attached") has three EQUAL side lengths.   Therefore, the triangle shown is an "equilateral triangle" and has 3 (three) equal angles, as well.

All triangles by, definition, have 3 (three) angles that add up to "180° " .

Since each of the 3 (three) angles is equal;  and the three angles are:  

  "∠A" , "∠B" , and "∠C" ;

We can find the measure of "∠A" ; denoted as:  "m∠A" ; as follows:
______________________________________________________
    m∠A  = 180° ÷ 3 = 60° .
______________________________________________________
The answer is:  " 60° " .
______________________________________________________
   m∠A = " 60° " .
______________________________________________________
7 0
3 years ago
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