Indefinite Integrals — Maths STD 12 Science — Question

$=\int\bigg(\frac{8\frac{\cos\text{x}}{\sin\text{x}}+1}{\frac{3\cos\text{x}}{\sin\text{x}}+2}\bigg)\text{dx}$

$=\int\Big(\frac{8\cos\text{x}+\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$

Now, Let $8\cos\text{x}+\sin\text{x}=\text{A}(3\cos\text{x}+2\sin\text{x})+\text{B}(-3\sin\text{x}+2\cos\text{x})\ \dots(1)$

$\Rightarrow8\cos\text{x}+\sin\text{x}+\sin\text{x}=3\text{A}\cos\text{x}+2\text{A}\sin\text{x}-3\text{B}\sin\text{x}+2\text{B}\cos\text{x}$

$\Rightarrow8\cos\text{x}+\sin\text{x}-(3\text{A}+2\text{B})\cos\text{x}+(2\text{A}-3\text{B})\sin\text{x}$

Equating the coefficient of like terms we get,

$2\text{A}-3\text{B}=1\ \dots(2)$

$3\text{A}+2\text{B}=8\ \dots(3)$

Solving eq (2) and eq (3) we get,

$\text{A}=2,\text{B}=1$

Thus, by substracting the value of A and B in eq (1) we get,

$\text{I}=\int\Big[\frac{2(3\cos\text{x}+2\sin\text{x})+1(-3\sin\text{x}+2\cos\text{x})}{(3\cos\text{x}+2\sin\text{x})}\Big]\text{dx}$

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

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

Putting $3\cos\text{x}+2\sin\text{x}=\text{t}$

$\Rightarrow(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$

$\therefore\text{I}=2\int\text{dx}+\int\frac{1}{\text{t}}\text{dt}$

$=2\text{x}+\ln|\text{t}|+\text{C}$

$=2\text{x}+\ln|3\cos\text{x}+2\sin\text{x}|+\text{C}$