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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following intregals:
$\int\frac{1}{(\sin\text{x}-2\cos\text{x})(2\sin\text{x}-\cos\text{x})}\ \text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{(\sin\text{x}-2\cos\text{x})(2\sin\text{x}-\cos\text{x})}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\Rightarrow\text{I}=\int\frac{\sec^2\text{x}}{\Big(\frac{\sin\text{x}-2\cos\text{x}}{\cos\text{x}}\Big)\times\Big(\frac{2\sin\text{x}+\cos\text{x}}{\cos\text{x}}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{(\tan^2\text{x}-2)(2\tan\text{x}+1)}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}-2)(2\text{t}-1 0)}$
$=\int\frac{\text{dt}}{2\text{t}^2+\text{t}-4\text{t}-2}$
$=\int\frac{\text{dt}}{2\text{t}^2-3\text{t}-2}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\frac{3}{2}\text{t}-1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\frac{3}{2}\text{t}+\Big(\frac{3}{4}\Big)^2-\Big(\frac{3}{4}\Big)^2-1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\text{t}-\frac{3}{4}\Big)-\frac{9}{16}-1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\text{t}-\frac{3}{4}\Big)-\Big(\frac{5}{4}\Big)^2}$
$=\frac{1}{2}\times\frac{1}{2\times\frac{5}{4}}\log\Bigg|\frac{\text{t}-\frac{3}{4}-\frac{5}{4}}{\text{t}-\frac{3}{4}+\frac{5}{4}}\Bigg|+\text{C}$
$=\frac{1}{5}\ln\Bigg|\frac{\text{t}-2}{\text{t}+\frac{1}{2}}\Bigg|+\text{C}$
$=\frac{1}{5}\ln\Big|\frac{\text{t}-2}{2\text{t}+1}\Big|+\frac{1}{5}\int(2)+\text{C}$
$=\frac{1}{5}\ln\Big|\frac{\text{t}-2}{2\text{t}+1}\Big|+\text{C}$ where $\text{C}=\text{C}+\frac{1}{5}\int(2)$
$=\frac{1}{5}\ln\Big|\frac{\tan\text{x}-2}{2\tan\text{x}+1}\Big|+\text{C}$

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