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# ∫ 1 13 + 3 Cos X + 4 Sin X D X - Mathematics

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Sum
$\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx$
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#### Solution

$\text{ Let I }= \int \frac{1}{13 + 3 \cos x + 4 \sin x}dx$
$\text{ Putting cos x }= \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \text{ and sin x }= \frac{2\tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)}$
$\therefore I = \int \frac{1}{13 + 3 \left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) + 4 \times 2\frac{\tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)}}dx$
$= \int \frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{13\left( 1 + \tan^2 \frac{x}{2} \right) + 3 - 3 \tan^2 \frac{x}{2} + 8 \tan \left( \frac{x}{2} \right)} dx$
$= \int \frac{\sec^2 \frac{x}{2}}{13 \tan^2 \frac{x}{2} - 3 \tan^2 \frac{x}{2} + 16 + 8 \tan \left( \frac{x}{2} \right)}dx$
$= \int \frac{\sec^2 \frac{x}{2}}{10 \tan^2 \left( \frac{x}{2} \right) + 8 \tan \left( \frac{x}{2} \right) + 16}dx$
$\text{ Let tan} \left( \frac{x}{2} \right) = t$
$\Rightarrow \frac{1}{2} \text{ sec}^2 \left( \frac{x}{2} \right)dx = dt$
$\Rightarrow \text{ sec}^2 \left( \frac{x}{2} \right)dx = 2dt$
$\therefore I = \int \frac{2 dt}{10 t^2 + 8t + 16}$
$= \int \frac{dt}{5 t^2 + 4t + 8}$
$= \frac{1}{5} \int \frac{dt}{t^2 + \frac{4}{5}t + \frac{8}{5}}$
$= \frac{1}{5}\int \frac{dt}{t^2 + \frac{4}{5}t + \left( \frac{2}{5} \right)^2 - \left( \frac{2}{5} \right)^2 + \frac{8}{5}}$

$= \frac{1}{5}\int \frac{dt}{\left( t + \frac{2}{5} \right)^2 - \frac{4}{25} + \frac{8}{5}}$
$= \frac{1}{5}\int \frac{dt}{\left( t + \frac{2}{5} \right)^2 + \frac{- 4 + 40}{25}}$
$= \frac{1}{5}\int \frac{dt}{\left( t + \frac{2}{5} \right)^2 + \left( \frac{6}{5} \right)^2}$
$= \frac{1}{5} \times \frac{5}{6} \tan^{- 1} \left( \frac{t + \frac{2}{5}}{\frac{6}{5}} \right) + C$
$= \frac{1}{6} \tan^{- 1} \left( \frac{5t + 2}{6} \right) + C$
$= \frac{1}{6} \tan^{- 1} \left( \frac{5 \tan \frac{x}{2} + 2}{6} \right) + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.23 | Q 7 | Page 117

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