Question
Evaluate the following integrals:$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$
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Answer
Let $\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$ $\therefore\ \text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-(1-\sin^2\text{x})-4\sin\text{x}}\text{ dx}$ $\Rightarrow\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-1+\sin^2\text{x}-4\sin\text{x}}\text{ dx}$ Substitute $\sin\text{x}=\text{t}$ $\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$ Thus, $\text{I}=\int\frac{(3\text{t}-2)}{4+\text{t}^2-4\text{t}}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{\text{t}^2-4\text{t}+4}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ Now let us separate the integrand into the simplest form using partial fractions. $\frac{(3\text{t}-2)}{(\text{t}-2)^2}=\frac{\text{A}}{(\text{t}-2)}+\frac{\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{A}(\text{t}-2)+\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{At}-2\text{A}+\text{B}}{(\text{t}-2)^2}$ $\Rightarrow3\text{t}-2=\text{At}-2\text{A}+\text{B}$ Comparing the coefficients, we have, $\text{A}=3$and
$-2\text{A}+\text{B}=-2$ Substituting the value of A = 3 in the above equation, we have, $\Rightarrow-2\times3+\text{B}=-2$ $\Rightarrow-6+\text{B}=-2$ $\Rightarrow\text{B}=6-2$ $\Rightarrow\text{B}=4$Thus, $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ becomes,
$\text{I}=\int\frac{3}{(\text{t}-2)^2}\text{ dt}+\int\frac{4}{(\text{t}-2)^2}\text{ dt}$ $=3\log|\text{t}-2|-4\Big(\frac{1}{\text{t}-2}\Big)+\text{C}$ $=3\log|2-\text{t}|+4\Big(\frac{1}{2-\text{t}}\Big)+\text{C}$ Now, substituting $\text{t}=\sin\text{x},$ we have, $=3\log|2-\sin\text{x}|+4\Big(\frac{1}{2-\sin\text{x}}\Big)+\text{C}$
$-2\text{A}+\text{B}=-2$ Substituting the value of A = 3 in the above equation, we have, $\Rightarrow-2\times3+\text{B}=-2$ $\Rightarrow-6+\text{B}=-2$ $\Rightarrow\text{B}=6-2$ $\Rightarrow\text{B}=4$Thus, $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ becomes,
$\text{I}=\int\frac{3}{(\text{t}-2)^2}\text{ dt}+\int\frac{4}{(\text{t}-2)^2}\text{ dt}$ $=3\log|\text{t}-2|-4\Big(\frac{1}{\text{t}-2}\Big)+\text{C}$ $=3\log|2-\text{t}|+4\Big(\frac{1}{2-\text{t}}\Big)+\text{C}$ Now, substituting $\text{t}=\sin\text{x},$ we have, $=3\log|2-\sin\text{x}|+4\Big(\frac{1}{2-\sin\text{x}}\Big)+\text{C}$
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