Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To evaluate the integral
[tex]\[ \int \frac{\cos x}{6 \sin ^2 x + 7 \sin x} \, dx, \][/tex]
we follow these steps:
1. Substitution:
Let [tex]\( u = \sin x \)[/tex]. Then, [tex]\( du = \cos x \, dx \)[/tex]. This simplifies our integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int \frac{\cos x}{6 \sin^2 x + 7 \sin x} \, dx = \int \frac{1}{6u^2 + 7u} \, du. \][/tex]
2. Partial Fraction Decomposition:
We decompose the integrand [tex]\( \frac{1}{6u^2 + 7u} \)[/tex]:
[tex]\[ \frac{1}{6u^2 + 7u} = \frac{1}{u(6u + 7)}. \][/tex]
This can be written using partial fractions:
[tex]\[ \frac{1}{u(6u + 7)} = \frac{A}{u} + \frac{B}{6u + 7}, \][/tex]
where [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are constants to be determined. To find [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we solve the equation:
[tex]\[ 1 = A(6u + 7) + Bu. \][/tex]
Setting [tex]\( u = 0 \)[/tex], we find:
[tex]\[ 1 = 7A \implies A = \frac{1}{7}. \][/tex]
To find [tex]\( B \)[/tex], set [tex]\( u = -\frac{7}{6} \)[/tex]:
[tex]\[ 1 = \frac{1}{7}(-\frac{7}{6} \cdot 6 + 7) + B(-\frac{7}{6}) \implies 1 = \frac{1}{7} \cdot 0 - \frac{7}{6}B \implies B = -\frac{6}{7}. \][/tex]
Therefore, the partial fractions decomposition is:
[tex]\[ \frac{1}{u(6u + 7)} = \frac{1}{7u} - \frac{6}{7(6u + 7)}. \][/tex]
3. Integration:
Now, we integrate each term separately:
[tex]\[ \int \left( \frac{1}{7u} - \frac{6}{7(6u + 7)} \right) \, du = \frac{1}{7} \int \frac{1}{u} \, du - \frac{6}{7} \int \frac{1}{6u + 7} \, du. \][/tex]
These integrals are straightforward:
[tex]\[ \int \frac{1}{u} \, du = \ln|u|, \][/tex]
and for the second integral, we use substitution [tex]\( v = 6u + 7 \)[/tex], then [tex]\( dv = 6 \, du \)[/tex], so [tex]\( du = \frac{dv}{6} \)[/tex]:
[tex]\[ \int \frac{1}{6u + 7} \, du = \frac{1}{6} \int \frac{1}{v} \, dv = \frac{1}{6} \ln|v| = \frac{1}{6} \ln|6u + 7|. \][/tex]
Putting it all together:
[tex]\[ \frac{1}{7} \int \frac{1}{u} \, du - \frac{6}{7} \int \frac{1}{6u + 7} \, du = \frac{1}{7} \ln|u| - \frac{6}{7} \cdot \frac{1}{6} \ln|6u + 7| = \frac{1}{7} \ln|u| - \frac{1}{7} \ln|6u + 7|. \][/tex]
4. Back Substitute:
Finally, revert back from [tex]\( u = \sin x \)[/tex]:
[tex]\[ \frac{1}{7} \ln|\sin x| - \frac{1}{7} \ln|6\sin x + 7|. \][/tex]
Combine the logarithms:
[tex]\[ \frac{1}{7} \left( \ln|\sin x| - \ln|6 \sin x + 7| \right) = \frac{1}{7} \ln \left| \frac{\sin x}{6 \sin x + 7} \right|. \][/tex]
So, the evaluated integral is:
[tex]\[ \int \frac{\cos x}{6 \sin^2 x + 7 \sin x} \, dx = -\frac{\ln(6 \sin x + 7)}{7} + \frac{\ln(\sin x)}{7} + C, \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
[tex]\[ \int \frac{\cos x}{6 \sin ^2 x + 7 \sin x} \, dx, \][/tex]
we follow these steps:
1. Substitution:
Let [tex]\( u = \sin x \)[/tex]. Then, [tex]\( du = \cos x \, dx \)[/tex]. This simplifies our integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int \frac{\cos x}{6 \sin^2 x + 7 \sin x} \, dx = \int \frac{1}{6u^2 + 7u} \, du. \][/tex]
2. Partial Fraction Decomposition:
We decompose the integrand [tex]\( \frac{1}{6u^2 + 7u} \)[/tex]:
[tex]\[ \frac{1}{6u^2 + 7u} = \frac{1}{u(6u + 7)}. \][/tex]
This can be written using partial fractions:
[tex]\[ \frac{1}{u(6u + 7)} = \frac{A}{u} + \frac{B}{6u + 7}, \][/tex]
where [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are constants to be determined. To find [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we solve the equation:
[tex]\[ 1 = A(6u + 7) + Bu. \][/tex]
Setting [tex]\( u = 0 \)[/tex], we find:
[tex]\[ 1 = 7A \implies A = \frac{1}{7}. \][/tex]
To find [tex]\( B \)[/tex], set [tex]\( u = -\frac{7}{6} \)[/tex]:
[tex]\[ 1 = \frac{1}{7}(-\frac{7}{6} \cdot 6 + 7) + B(-\frac{7}{6}) \implies 1 = \frac{1}{7} \cdot 0 - \frac{7}{6}B \implies B = -\frac{6}{7}. \][/tex]
Therefore, the partial fractions decomposition is:
[tex]\[ \frac{1}{u(6u + 7)} = \frac{1}{7u} - \frac{6}{7(6u + 7)}. \][/tex]
3. Integration:
Now, we integrate each term separately:
[tex]\[ \int \left( \frac{1}{7u} - \frac{6}{7(6u + 7)} \right) \, du = \frac{1}{7} \int \frac{1}{u} \, du - \frac{6}{7} \int \frac{1}{6u + 7} \, du. \][/tex]
These integrals are straightforward:
[tex]\[ \int \frac{1}{u} \, du = \ln|u|, \][/tex]
and for the second integral, we use substitution [tex]\( v = 6u + 7 \)[/tex], then [tex]\( dv = 6 \, du \)[/tex], so [tex]\( du = \frac{dv}{6} \)[/tex]:
[tex]\[ \int \frac{1}{6u + 7} \, du = \frac{1}{6} \int \frac{1}{v} \, dv = \frac{1}{6} \ln|v| = \frac{1}{6} \ln|6u + 7|. \][/tex]
Putting it all together:
[tex]\[ \frac{1}{7} \int \frac{1}{u} \, du - \frac{6}{7} \int \frac{1}{6u + 7} \, du = \frac{1}{7} \ln|u| - \frac{6}{7} \cdot \frac{1}{6} \ln|6u + 7| = \frac{1}{7} \ln|u| - \frac{1}{7} \ln|6u + 7|. \][/tex]
4. Back Substitute:
Finally, revert back from [tex]\( u = \sin x \)[/tex]:
[tex]\[ \frac{1}{7} \ln|\sin x| - \frac{1}{7} \ln|6\sin x + 7|. \][/tex]
Combine the logarithms:
[tex]\[ \frac{1}{7} \left( \ln|\sin x| - \ln|6 \sin x + 7| \right) = \frac{1}{7} \ln \left| \frac{\sin x}{6 \sin x + 7} \right|. \][/tex]
So, the evaluated integral is:
[tex]\[ \int \frac{\cos x}{6 \sin^2 x + 7 \sin x} \, dx = -\frac{\ln(6 \sin x + 7)}{7} + \frac{\ln(\sin x)}{7} + C, \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
