Evalute the following integrals: 2x 3x 5x dx - Vidyadip

Question

Evalute the following integrals:
$\int\tan2\text{x}\tan3\text{x}\tan5\text{x dx}$

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Answer

Let $\text{I}=\int1+\tan\text{x}\tan(\text{x}+\theta)\text{dx}$
$=\int1+\tan\text{x}\Big(\frac{\tan\text{x}+\tan\theta}{1-\tan\text{x}\tan\theta}\Big)\text{dx}$
$=\int\frac{1+\tan^2\text{x}}{1-\tan\text{x}\tan\theta}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{1-\tan\text{x}\tan\theta}$
Putting $\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
$\therefore\text{I}\approx\int\frac{1}{1-\text{t}\tan\theta}\text{dt}$
$=\frac{-1}{\tan\theta}\text{ ln}|1-\text{t}\tan\theta|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=-\cot\theta\text{ ln}|1-\tan\text{ x }\tan\theta|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{1}{1-\tan\text{ x }\tan\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{ x }\cos\theta}{\cos\text{x}\cos\theta-\sin\text{x}\sin\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C }\big[\text{Let C}'=\text{C}+\cot\theta\text{ ln}\cos\theta\big]$

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