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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate the following intregals:
$\int\frac{1}{13+3\cos\text{x}+4\sin\text{x}}\ \text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{13+3\cos\text{x}+4\sin\text{x}}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\sin\text{x}=\frac{2\tan\Big(\frac{\text{x}}{2}\Big)}{1+\tan^2\Big(\frac{\text{x}}{2}\Big)}$
$\therefore\text{I}=\int\frac{1}{13+3\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+4\times2\frac{\tan\Big(\frac{\text{x}}{2}\Big)}{1+\tan^2\Big(\frac{\text{x}}{2}\Big)}}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{13\Big(1+\tan^2\frac{\text{x}}{2}\Big)+3-3\tan^2\frac{\text{x}}{2}+16+8\tan\Big(\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{13\tan^2\frac{\text{x}}{2}-3\tan^2\frac{\text{x}}{2}+16+8\tan\Big(\frac{\text{x}}{2}\Big)}\ \text{dx}$
$==\int\frac{\sec^2\frac{\text{x}}{2}}{10\tan^2\Big(\frac{\text{x}}{2}\Big)+8\tan\Big(\frac{\text{x}}{2}\Big)+16}\ \text{dx}$
Let $\tan\Big(\frac{\text{x}}{2}\Big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\text{I}=\int\frac{2\text{dt}}{10\text{t}^2+8\text{t}+16}$
$=\int\frac{\text{dt}}{5\text{t}^2+4\text{t}+8}$
$=\frac{1}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{4}{5}\text{t}+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{4}{5}\text{t}+\Big(\frac{2}{5}\Big)^2-\Big(\frac{2}{5}\Big)^2+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)^2-\frac{4}{25}+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)^2+\frac{-4+40}{25}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)+\Big(\frac{6}{5}\Big)^2}$
$=\frac{1}{5}\times\frac{5}{6}\tan^{-1}\Bigg(\frac{\text{t}+\frac{2}{5}}{\frac{6}{5}}\Bigg)+\text{C}$
$=\frac{1}{6}\tan^{-1}\Big(\frac{5\text{t}+2}{6}\Big)+\text{C}$
$=\frac{1}{6}\tan^{-1}\bigg(\frac{5\tan\frac{\text{x}}{2}+2}{6}\bigg)+\text{C}$

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