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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Find the integrals of the functions in Exercises:
$\frac{\cos2\text{x}-\cos2\alpha}{\cos\text{x}-\cos\alpha}$
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Answer

$\frac{\cos2\text{x}-\cos2\alpha}{\cos\text{x}-\cos\alpha}=\frac{-2\sin\frac{2\text{x}+2\alpha}{2}\sin\frac{2\text{x}-2\alpha}{2}}{-2\sin\frac{\text{x}+\alpha}{2}\sin\frac{\text{x}-\alpha}{2}}$$\ \ \ \ \ \ \ \bigg[\cos\text{C}-\cos\text{D}=-2\sin\frac{\text{C}+\text{D}}{2}\sin\frac{\text{C}-\text{D}}{2}\bigg]$
$=\frac{\sin(\text{x}+\alpha)\sin(\text{x}-\alpha)}{\sin\bigg(\frac{\text{x}+\alpha}{2}\bigg)\sin\bigg(\frac{\text{x}-\alpha}{2}\bigg)}$
$=\frac{\bigg[2\sin\bigg(\frac{\text{x}+\alpha}{2}\bigg)\cos\bigg(\frac{\text{x}+\alpha}{2}\bigg)\bigg]\bigg[2\sin\bigg(\frac{\text{x}-\alpha}{2}\bigg)\cos\bigg(\frac{\text{x}-\alpha}{2}\bigg)\bigg]}{\sin\bigg(\frac{\text{x}+\alpha}{2}\bigg)\sin\bigg(\frac{\text{x}-\alpha}{2}\bigg)}$
$=4\cos\bigg(\frac{\text{x}+\alpha}{2}\bigg)\cos\bigg(\frac{\text{x}-\alpha}{2}\bigg)$
$=2\bigg[\cos\bigg(\frac{\text{x}+\alpha}{2}+\frac{\text{x}-\alpha}{2}\bigg)+\cos\frac{\text{x}+\alpha}{2}-\frac{\text{x}-\alpha}{2}\bigg]$
$=2\big[\cos\text{(x)}+\cos\alpha\big]$
$=2\cos\text{x}+2\cos\alpha$
$\therefore\int\frac{\cos2\text{x}-\cos2\alpha}{\cos\text{x}-\cos\alpha}\text{dx}=\int2\cos\text{x}+2\cos\alpha$
$=2\big[\sin\text{x}+\text{x}\cos\alpha\big]+\text{C}$

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