∫ 1 4 Cos X − 1 D X - Mathematics

Sum
$\int\frac{1}{4 \cos x - 1} \text{ dx }$

Solution

$\text{ Let I} = \int \frac{1}{4 \cos x - 1}dx$
$\text{ Putting cos x }= \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}$
$\Rightarrow I = \int \frac{1}{4\left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) - 1}dx$
$= \int \frac{1}{\frac{4\left( 1 - \tan^2 \frac{x}{2} \right) - \left( 1 + \tan^2 \frac{x}{2} \right)}{\left( 1 + \tan^2 \frac{x}{2} \right)}}$
$= \int \frac{\left( 1 + \tan^2 \frac{x}{2} \right)dx}{4 - 4 \tan^2 \left( \frac{x}{2} \right) - 1 - \tan^2 \left( \frac{x}{2} \right)}$
$= \int \frac{\sec^2 \left( \frac{x}{2} \right) dx}{3 - 5 \tan^2 \left( \frac{x}{2} \right)}$
$\text{ Let tan } \left( \frac{x}{2} \right) = t$
$\Rightarrow \frac{1}{2} \sec^2 \left( \frac{x}{2} \right)\text{ dx }= dt$
$\Rightarrow \sec^2 \left( \frac{x}{2} \right)dx = 2dt$
$\therefore I = 2 \int \frac{dt}{3 - 5 t^2}$
$= \frac{2}{5} \int \frac{dt}{\frac{3}{5} - t^2}$
$= \frac{2}{5} \int \frac{dt}{\left( \frac{\sqrt{3}}{\sqrt{5}} \right)^2 - t^2}$
$= \frac{2}{5} \times \frac{\sqrt{5}}{2\sqrt{3}}\text{ In }\left| \frac{\frac{\sqrt{3}}{\sqrt{5}} + t}{\frac{\sqrt{3}}{\sqrt{5}} - t} \right| + C$
$= \frac{1}{\sqrt{15}}\text{ ln } \left| \frac{\sqrt{3} + \sqrt{5} t}{\sqrt{3} - \sqrt{5} t} \right| + C$
$= \frac{1}{\sqrt{15}}\text{ ln }\left| \frac{\sqrt{3} + \sqrt{5} \tan \left( \frac{x}{2} \right)}{\sqrt{3} - \sqrt{5} \tan \left( \frac{x}{2} \right)} \right| + C$

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.23 | Q 4 | Page 117