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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Integrate the function in Exercise:$\frac{1}{\sqrt{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}}$
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Answer

$\frac{1}{\sqrt{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}}=\frac{1}{\sqrt{\sin^{3}\text{x}(\sin\text{x}\cos\text{a}+\cos\text{x}\sin\text{a)}}}$
$=\frac{1}{\sqrt{\sin^{4}\text{x}\cos\text{a}+\sin^{3}\text{x}\cos\text{x}\sin\text{a}}}$
$=\frac{1}{\sin^{2}\text{x}\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}}$
$=\frac{\text{cosec}^{2}\text{x}}{\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}}$
$\text{Let}\cos\text{a}+\cot\text{x}\sin\text{a}=\text{t}\Rightarrow-\text{cosec}^{2}\text{x}\sin\text{a}\ \text{dx}=\text{dt}$
$\therefore\int\frac{1}{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}\text{dx}-\frac{\text{cosec}^{2}\text{x}}{\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}}\text{dx}$
$=\frac{-1}{\sin\text{a}}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\frac{-1}{\sin\text{a}}\big[2\sqrt{\text{t}}\big]+\text{C}$
$=\frac{-1}{\sin\text{a}}\big[2\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}\big]+\text{C}$
$=\frac{-2}{\sin\text{a}}\sqrt{\cos\text{a}+\frac{\cos\text{x}\sin\text{a}}{\sin\text{x}}}+\text{C}$
$=\frac{-2}{\sin\text{a}}\sqrt{\frac{\sin\text{x}\cos\text{a}+\cos\text{x}\sin\text{a}}{\sin\text{x}}}+\text{C}$
$=-\frac{2}{\sin\text{a}}\sqrt{\frac{\sin(\text{x}+\text{a)}}{\sin\text{x}}}+\text{C}$

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