Evaluate the following integrals: (3 -2) 5- ^2x-4 dx

Question

Evaluate the following integrals:

$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$

✓

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,

$=3\log|2-\sin\text{x}|+4\Big(\frac{1}{2-\sin\text{x}}\Big)+\text{C}$

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