Answer :
To solve the integral
[tex]\[ \int_2^3 \frac{x}{\sqrt{169-x^2}} \, dx, \][/tex]
we can follow these steps:
1. Identify the integrand: The integrand is [tex]\(\frac{x}{\sqrt{169 - x^2}}\)[/tex]. It suggests that a trigonometric substitution could be useful, or we might recognize a pattern of a derivative of an inverse trigonometric function.
2. Consider trigonometric substitution: Here, we recognize that [tex]\(169\)[/tex] is [tex]\(13^2\)[/tex], so we can use the substitution [tex]\( x = 13 \sin \theta \)[/tex], which implies that [tex]\(dx = 13 \cos \theta \, d\theta\)[/tex]. Also, [tex]\(13^2 - x^2 = 169 - 169 \sin^2 \theta = 169 \cos^2 \theta \)[/tex]. Hence, [tex]\( \sqrt{169 - x^2} = 13 \cos \theta \)[/tex].
3. Apply the substitution: Substitute [tex]\(x = 13 \sin \theta \)[/tex] and [tex]\(dx = 13 \cos \theta \, d\theta\)[/tex] into the integral.
[tex]\[ \int_2^3 \frac{x}{\sqrt{169-x^2}} \, dx = \int_2^3 \frac{13 \sin \theta}{13 \cos \theta} \cdot 13 \cos \theta \, d\theta = \int_2^3 \sin \theta \cdot 13 \, d\theta \][/tex]
4. Adjust the limits of integration: When [tex]\(x = 2\)[/tex], [tex]\(2 = 13 \sin \theta\)[/tex] thus [tex]\(\sin \theta = \frac{2}{13}\)[/tex], and when [tex]\(x = 3\)[/tex], [tex]\(3 = 13 \sin \theta\)[/tex] thus [tex]\(\sin \theta = \frac{3}{13}\)[/tex].
5. Evaluate the new integral: The integral in terms of [tex]\(\theta\)[/tex] is:
[tex]\[ \int_{\arcsin(\frac{2}{13})}^{\arcsin(\frac{3}{13})} 13 \sin \theta \, d\theta \][/tex]
6. Solve the integral: Using simple antiderivatives for the basic trigonometric function:
[tex]\[ \int 13 \sin \theta \, d\theta = -13 \cos \theta \][/tex]
7. Substitute back and evaluate at the bounds:
[tex]\[ \left[ -13 \cos \theta \right]_{\arcsin(\frac{2}{13})}^{\arcsin(\frac{3}{13})} \][/tex]
8. Calculate the cosine values of the bounds:
[tex]\[ -13 \cos(\arcsin(\frac{3}{13})) - (-13 \cos(\arcsin(\frac{2}{13}))) \][/tex]
9. Simplify using the relationships [tex]\( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} \)[/tex]:
For [tex]\(\theta = \arcsin(\frac{2}{13})\)[/tex]:
[tex]\[ \cos(\arcsin(\frac{2}{13})) = \sqrt{1 - \left(\frac{2}{13}\right)^2} = \sqrt{\frac{169 - 4}{169}} = \sqrt{\frac{165}{169}} = \frac{\sqrt{165}}{13} \][/tex]
For [tex]\(\theta = \arcsin(\frac{3}{13})\)[/tex]:
[tex]\[ \cos(\arcsin(\frac{3}{13})) = \sqrt{1 - \left(\frac{3}{13}\right)^2} = \sqrt{\frac{169 - 9}{169}} = \sqrt{\frac{160}{169}} = \frac{\sqrt{160}}{13} \][/tex]
10. Substitute these values in:
[tex]\[ -13 \cdot \frac{\sqrt{160}}{13} - \left( -13 \cdot \frac{\sqrt{165}}{13} \right) \][/tex]
11. Combine and simplify:
[tex]\[ -\sqrt{160} + \sqrt{165} \][/tex]
Therefore, the value of the integral is:
[tex]\[ \boxed{-4\sqrt{10} + \sqrt{165}} \][/tex]
[tex]\[ \int_2^3 \frac{x}{\sqrt{169-x^2}} \, dx, \][/tex]
we can follow these steps:
1. Identify the integrand: The integrand is [tex]\(\frac{x}{\sqrt{169 - x^2}}\)[/tex]. It suggests that a trigonometric substitution could be useful, or we might recognize a pattern of a derivative of an inverse trigonometric function.
2. Consider trigonometric substitution: Here, we recognize that [tex]\(169\)[/tex] is [tex]\(13^2\)[/tex], so we can use the substitution [tex]\( x = 13 \sin \theta \)[/tex], which implies that [tex]\(dx = 13 \cos \theta \, d\theta\)[/tex]. Also, [tex]\(13^2 - x^2 = 169 - 169 \sin^2 \theta = 169 \cos^2 \theta \)[/tex]. Hence, [tex]\( \sqrt{169 - x^2} = 13 \cos \theta \)[/tex].
3. Apply the substitution: Substitute [tex]\(x = 13 \sin \theta \)[/tex] and [tex]\(dx = 13 \cos \theta \, d\theta\)[/tex] into the integral.
[tex]\[ \int_2^3 \frac{x}{\sqrt{169-x^2}} \, dx = \int_2^3 \frac{13 \sin \theta}{13 \cos \theta} \cdot 13 \cos \theta \, d\theta = \int_2^3 \sin \theta \cdot 13 \, d\theta \][/tex]
4. Adjust the limits of integration: When [tex]\(x = 2\)[/tex], [tex]\(2 = 13 \sin \theta\)[/tex] thus [tex]\(\sin \theta = \frac{2}{13}\)[/tex], and when [tex]\(x = 3\)[/tex], [tex]\(3 = 13 \sin \theta\)[/tex] thus [tex]\(\sin \theta = \frac{3}{13}\)[/tex].
5. Evaluate the new integral: The integral in terms of [tex]\(\theta\)[/tex] is:
[tex]\[ \int_{\arcsin(\frac{2}{13})}^{\arcsin(\frac{3}{13})} 13 \sin \theta \, d\theta \][/tex]
6. Solve the integral: Using simple antiderivatives for the basic trigonometric function:
[tex]\[ \int 13 \sin \theta \, d\theta = -13 \cos \theta \][/tex]
7. Substitute back and evaluate at the bounds:
[tex]\[ \left[ -13 \cos \theta \right]_{\arcsin(\frac{2}{13})}^{\arcsin(\frac{3}{13})} \][/tex]
8. Calculate the cosine values of the bounds:
[tex]\[ -13 \cos(\arcsin(\frac{3}{13})) - (-13 \cos(\arcsin(\frac{2}{13}))) \][/tex]
9. Simplify using the relationships [tex]\( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} \)[/tex]:
For [tex]\(\theta = \arcsin(\frac{2}{13})\)[/tex]:
[tex]\[ \cos(\arcsin(\frac{2}{13})) = \sqrt{1 - \left(\frac{2}{13}\right)^2} = \sqrt{\frac{169 - 4}{169}} = \sqrt{\frac{165}{169}} = \frac{\sqrt{165}}{13} \][/tex]
For [tex]\(\theta = \arcsin(\frac{3}{13})\)[/tex]:
[tex]\[ \cos(\arcsin(\frac{3}{13})) = \sqrt{1 - \left(\frac{3}{13}\right)^2} = \sqrt{\frac{169 - 9}{169}} = \sqrt{\frac{160}{169}} = \frac{\sqrt{160}}{13} \][/tex]
10. Substitute these values in:
[tex]\[ -13 \cdot \frac{\sqrt{160}}{13} - \left( -13 \cdot \frac{\sqrt{165}}{13} \right) \][/tex]
11. Combine and simplify:
[tex]\[ -\sqrt{160} + \sqrt{165} \][/tex]
Therefore, the value of the integral is:
[tex]\[ \boxed{-4\sqrt{10} + \sqrt{165}} \][/tex]
