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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals3 Marks
Question
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{1-\cos\text{x}}\text{dx}\text{ or }\int\frac{\cot\text{x}}{\text{cosec x}-\cot\text{x}}\text{dx}$
✓

Answer

$\int\frac{\cot\text{x}}{\text{cosec x}-\cot\text{x}}\text{dx}$
$=\int\frac{\frac{\cos\text{x}}{\sin\text{x}}}{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{1-\cos\text{x}}\Big)\times\frac{(1+\cos\text{x})}{(1+\cos\text{x})}\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{1-\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\big[(\cot\text{x}\text{ cosec x})+\cot^2\text{x}\big]\text{dx}$
$=\int\big[\cot\text{x}\text{ cosec x}+\text{cosec}^2\text{x}-1\big]\text{dx}$
$=-\text{cosec x}-\cot\text{x}-\text{x}+\text{C}$

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