int {frac{{left( {3sin phi - 2} right)cos ph | vedclass

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(D) Let $\sin \phi = t$,then $\cos \phi \, d\phi = dt$. Substituting this into the integral: $I = \int \frac{(3t - 2) dt}{5 - (1 - t^2) - 4t} = \int \

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$3\log \left( {2 - \sin \phi } \right) + \frac{4}{{\left( {2 - \sin \phi } \right)}} + C$

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(D) Let $\sin \phi = t$,then $\cos \phi \, d\phi = dt$. Substituting this into the integral: $I = \int \frac{(3t - 2) dt}{5 - (1 - t^2) - 4t} = \int \frac{3t - 2}{t^2 - 4t + 4} dt = \int \frac{3t - 2}{(t - 2)^2} dt$. Using partial fractions: $\frac{3t - 2}{(t - 2)^2} = \frac{A}{t - 2} + \frac{B}{(t - 2)^2}$. $3t - 2 = A(t - 2) + B$. Comparing coefficients,$A = 3$ and $B = 4$. Thus,$I = \int \frac{3}{t - 2} dt + \int \frac{4}{(t - 2)^2} dt$. $I = 3 \log |t - 2| - \frac{4}{t - 2} + C$. Since $t = \sin \phi$,we have $I = 3 \log |\sin \phi - 2| - \frac{4}{\sin \phi - 2} + C$. This simplifies to $I = 3 \log (2 - \sin \phi) + \frac{4}{2 - \sin \phi} + C$.

$\int {\frac{{\left( {3\sin \phi - 2} \right)\cos \phi }}{{5 - {{\cos }^2}\phi - 4\sin \phi }}\,} d\phi$ is equal to

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