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∫ 1 Sin X + √ 3 Cos X D X - Mathematics

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Sum
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx  }\]
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Solution

\[\text{ Let I }= \int \frac{1}{\sin x + \sqrt{3} \cos x}dx\]
\[\text{ Putting  sin x} = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \text{ and }\cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \Rightarrow I = \int \frac{1}{\frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \sqrt{3}\frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]
\[ = \int \frac{1 + \tan^2 \frac{x}{2}}{2 \tan \frac{x}{2} + \sqrt{3} - \sqrt{3} \tan^2 \frac{x}{2}}dx\]
\[ = \int\frac{\sec^2 \frac{x}{2}}{- \sqrt{3} \tan^2 \frac{x}{2} + 2 \tan \frac{x}{2} + \sqrt{3}}dx\]

\[\text{ Let }\tan \frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2} \text{ sec}^2 \frac{x}{2}dx = dt\]
\[ \Rightarrow \sec^2 \frac{x}{2}dx = 2dt\]
\[ \therefore I = 2\int \frac{dt}{- \sqrt{3} t^2 + 2t + \sqrt{3}}\]
\[ = - \frac{2}{\sqrt{3}}\int \frac{dt}{t^2 - \frac{2}{\sqrt{3}}t - 1}\]
\[ = - \frac{2}{\sqrt{3}}\int\frac{dt}{t^2 - \frac{2}{\sqrt{3}}t + \left( \frac{1}{\sqrt{3}} \right)^2 - \left( \frac{1}{\sqrt{3}} \right)^2 - 1}\]
\[ = - \frac{2}{\sqrt{3}}\int \frac{dt}{\left( t - \frac{1}{\sqrt{3}} \right)^2 - \left( \frac{2}{\sqrt{3}} \right)^2}\]
\[ = - \frac{2}{\sqrt{3}} \times \frac{1}{2\frac{2}{\sqrt{3}}}\text{ log      }\left| \frac{t - \frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}}{t - \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}} \right| + C\]

\[= - \frac{1}{2}\text{ log }\left| \frac{t - \frac{3}{\sqrt{3}}}{t + \frac{1}{\sqrt{3}}} \right| + C\]
\[ = - \frac{1}{2}\text{ log }\left| \frac{\sqrt{3}t - 3}{\sqrt{3}t + 1} \right| + C\]
\[ = \frac{1}{2}\text{ log }\left| \frac{\sqrt{3}t + 1}{\sqrt{3}t - 3} \right| + C\]
\[ = \frac{1}{2}\text{ log }\left| \frac{\sqrt{3}\tan\frac{x}{2} + 1}{\sqrt{3}\tan\frac{x}{2} - 3} \right| + C\]
\[or, \frac{1}{2}\text{ log }\left| \frac{1 + \sqrt{3}\tan\frac{x}{2}}{3 - \sqrt{3}\tan\frac{x}{2}} \right| + C\]

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.23 | Q 12 | Page 117

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