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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.
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