For trigonometric substitution to solve the above integral, fill in the blanks below using the picture of the triangle given.

integral (x^3)/(sqrt(25+9x^2)) dx
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/ l side A
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side B

*** sorry hard to draw on the computer

side a=3x
side b=?
side c=?
3x/5=?
(3/5) dx=?
(sqrt(25+9x^2))/(5)=?

well, if side c is the hypotenuse, then

side c= sqrt(25+9x^2)
side a=3x
side b=5
3x/5 = tan C
3/5 dx= d(tanC)
Sqrt(25+9x^2)/5=secC

To apply trigonometric substitution to solve the given integral, we need to set up a right triangle using the given information. Let's start with the triangle:

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Recognizing that side A is equal to 3x, we can label it as follows:

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/____________|
3x

Now, for the remaining sides, we need to determine their lengths using the Pythagorean theorem. The Pythagorean theorem states that the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. In this case, side B will be one of the legs, and side C will be the hypotenuse.

Applying the theorem, we have:

Side A^2 + Side B^2 = Side C^2
(3x)^2 + Side B^2 = Side C^2
9x^2 + Side B^2 = Side C^2

To solve for Side B, we need to isolate it:

Side B^2 = Side C^2 - 9x^2
Side B = sqrt(Side C^2 - 9x^2)

Now, we need to determine the value of Side C. Looking at the integral provided, we can see that the numerator is x^3, and the denominator is sqrt(25+9x^2). By comparing this to the Pythagorean identity sin^2(theta) + cos^2(theta) = 1, we can make the substitution x = (5/3)t, which implies (dx) = (5/3)dt.

Now, let's substitute the values we have:

Side a = 3x
Side b = sqrt(Side C^2 - 9x^2) -> Using the substitution: Side b = sqrt(25t^2 - 9(5/3)^2t^2)
Side c = ?

To find Side c, we need to use the trigonometric identity that relates the sides of a right triangle to an angle theta:

sin(theta) = side opposite / hypotenuse

From the given integral, we can write the following equation:

sin(theta) = sqrt(25+9x^2) / 5

Substituting x = (5/3)t, we have:

sin(theta) = sqrt(25+9(25/9)t^2) / 5
sin(theta) = sqrt(25+25t^2) / 5
sin(theta) = sqrt(25(1+t^2)) / 5

Now, we can determine the value of Side c:

Side c = 5sin(theta)
Side c = 5(sqrt(1+t^2))

Thus, the filled-in values are:

Side a = 3x
Side b = sqrt(25t^2 - 9(5/3)^2t^2)
Side c = 5(sqrt(1+t^2))
3x/5 = t
(3/5)dx = dt
(sqrt(25+9x^2))/(5) = sqrt(1+t^2)

To solve the given integral using trigonometric substitution, we need to use the triangle given in the picture.

Let's identify the sides of the triangle:

side a = 3x
side b = ?
side c = ?

To find side b, we can use the Pythagorean theorem:

a^2 + b^2 = c^2

(3x)^2 + b^2 = c^2

9x^2 + b^2 = c^2

Now, let's find the value of 3x/5:

3x/5 = (3/5)x

Lastly, let's find the value of (3/5)dx:

To find this, we need to differentiate (3/5)x with respect to x:

d((3/5)x)/dx = 3/5

Therefore, (3/5)dx = 3/5 dx.

Lastly, let's find the value of (sqrt(25+9x^2))/(5):

Since side c is the hypotenuse, we can write:

c = sqrt(25 + 9x^2)

Therefore,

(sqrt(25 + 9x^2))/5 = c/5.

Now that we have filled in the blanks, we can proceed with the trigonometric substitution.