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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS3 Marks
Question
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
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Answer

Let $\text{I}=\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\sin^2\frac{\pi}{6}}\text{dx}$
$\big[\because \sin(\text{A}+\text{B})\sin(\text{A}-\text{B})=\sin^2\text{A}-\sin^2\text{B}\big]$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\frac{1}{4}}\text{dx}$
Putting $\sin^2\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow2\sin\text{x}\cos\text{ x dx}=\text{dt}$
$\Rightarrow\sin2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}\big|\sin^2\text{x}-\frac{1}{4}\big|+\text{C}\ \Big[\because\text{t}=\sin^2\text{x}-\frac{1}{4}\Big]$

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