Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
