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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following intregals:
$\int\frac{1}{5+7\cos\text{x}+\sin\text{x}}\ \text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{5+7\cos\text{x}+\sin\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\text{ and }\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
Now,
$\text{I}=\int\frac{1}{5+\frac{7\Big(1-\tan\frac{\text{x}}{2}\Big)}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}+\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+7-7\tan^2\frac{\text{x}}{2}+2\tan\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{-2\tan^2\frac{\text{x}}{2}+12+2\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\text{I}=\int\frac{2\text{dt}}{-2\text{t}^2+12+2\text{t}}$
$=-\int\frac{\text{dt}}{\text{t}^2-\text{t}-6}$
$=-\int\frac{\text{dt}}{\text{t}^2-2\text{t}\Big(\frac{1}{2}\Big)+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2-6}$
$=-\int\frac{\text{dt}}{\Big(\text{t}-\frac{1}{2}\Big)^2-\Big(\frac{5}{2}\Big)^2}$
$=-\frac{1}{2\Big(\frac{5}{2}\Big)}\log\Bigg|\frac{\text{t}-\frac{1}{2}-\frac{5}{2}}{\text{t}-\frac{1}{2}+\frac{5}{2}}\Bigg|+\text{C}$
$=-\frac{1}{5}\log\Big|\frac{\text{t}-3}{\text{t}+2}\Big|+\text{C}$
$\text{I}=\frac{1}{5}\log\Bigg|\frac{\tan\frac{\text{x}}{2}+2}{\tan\frac{\text{x}}{2}-3}\Bigg|+\text{C}$

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