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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following intregals:
$\int\frac{1}{\sin\text{x}-\sqrt{3}\cos\text{x}}\ \text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{\sin\text{x}-\sqrt{3}\cos\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\text{ and }\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\Rightarrow\text{I}=\int\frac{1}{\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}-\sqrt{3}\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}-\sqrt{3}+\sqrt{3}\tan^2\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{\sqrt{3}\tan^2\frac{\text{x}}{2}+2\tan\frac{\text{x}}{2}-\sqrt{3}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\Rightarrow\sec^2\frac{\text{x}}{2}\text{dx}=2\text{dt}$
$\therefore\text{I}=2\int\frac{\text{dt}}{\sqrt{3}\text{t}^2+2\text{t}-\sqrt{3}}$
$=\frac{2}{\sqrt{3}}\int\frac{\text{dt}}{\text{t}^2+\frac{2}{\sqrt{3}}\text{t}-1}$
$=2\int\frac{\text{dt}}{\text{t}^2+\frac{2}{\sqrt{3}}\text{t}+\Big(\frac{1}{\sqrt{3}}\Big)^2-1-\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=2\int\frac{\text{dt}}{\Big(\text{t}+\frac{1}{\sqrt{3}}\Big)^2-\Big(\frac{2}{\sqrt{3}}\Big)^2}$
$=\frac{2}{2\times\frac{2}{\sqrt{3}}}\log\Bigg|\frac{\text{t}+\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}}}{\text{t}+\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}}\Bigg|+\text{C}$
$=\frac{\sqrt{3}}{2}\log\Bigg|\frac{\tan\frac{\text{x}}{2}-\frac{1}{\sqrt{3}}}{\tan\frac{\text{x}}{2}+\frac{3}{\sqrt{3}}}\Bigg|+\text{C}$
$=\frac{\sqrt{3}}{2}\log\Bigg|\frac{\sqrt{3}\tan\frac{\text{x}}{2}-1}{\sqrt{3}\tan\frac{\text{x}}{2}+3}\Bigg|+\text{C}$

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