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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
MCQ
Choose the correct option from given four options:
$\frac{\text{dx}}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}$ is equal to:
  • A
    $\sin(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
  • B
    $\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\Big|+\text{C}$
  • ✓
    $\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
  • D
    $\sin(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
✓

Answer

Correct option: C.
$\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
Let $\text{I}=\frac{\text{dx}}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{b}-\text{a})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a}-\text{x}+\text{b})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin\big\{(\text{x}-\text{a})-(\text{x}+\text{b})\big\}}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})-\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}-\frac{\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a} )\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\cos(\text{x}-\text{b})}{\sin(\text{x}-​​​​\text{a})}-\frac{\cos(\text{x}-\text{a})}{\sin(\text{x}-\text{a} )}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\big[\log\sin|(\text{x}-\text{b})|-\log|\sin(\text{x}-\text{b})|\big]+\text{C}$

$=\text{cosec}(\text{b}-\text{a})\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$

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