Evaluate the following integrals: (3 -2) 13- ^2x-7 dx

Question

Evaluate the following integrals:

$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$

✓

Answer

$\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$

$=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-(1-\sin^2\text{x})-7\sin\text{x}}\text{ dx}$

$\big(\because\ \cos^2\text{x}=1-\sin^2\text{x}\big)$

$=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-7\sin\text{x}+12}\text{ dx}$

$=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-4\sin\text{x}-3\sin\text{x}+12}\text{ dx}$

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

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

Let $\sin\text{x}=\text{t}$

$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$

$\therefore\ \text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}\text{ dt}$

Using partial fraction, we get

$\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}=\frac{\text{A}}{(\text{t}-3)}+\frac{\text{B}}{(\text{t}-4)}$

$=\frac{\text{A}(\text{t}-4)+\text{B}(\text{t}-3)}{(\text{t}-3)(\text{t}-4)}$

$\Rightarrow3\text{t}-2=(\text{A}+\text{B})\text{t}-4\text{A}-3\text{B}$

Comparing coefficients, we get

A = -7 and B = 10

So, $\text{I}=-7\int\frac{1}{(\text{t}-3)}\text{ dt}+10\int\frac{1}{(\text{t}-4)}\text{ dt}$

$\Rightarrow\text{I}=-7\ln|\text{t}-3|+10\ln|\text{t}-4|+\text{C}$

$\therefore\ \text{I}=-7\ln|\sin\text{x}-3|+10\ln|\sin\text{x}-4|+\text{C}$

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