$\int t a n 2 x t a n 3 x t a n 5 x d x =$

\frac{1}{5} l o g | s e c 2 x | - \frac{1}{3} l o g | s e c 3 x | - \frac{1}{5} l o g | s e c x | + c

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\frac{1}{2} l o g | s e c 2 x | - \frac{1}{3} l o g | s e c 3 x | - \frac{1}{5} l o g | s e c 5 x | + c

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\frac{1}{5} l o g | s e c 5 x | - \frac{1}{2} l o g | s e c 2 x | - \frac{1}{3} l o g | s e c 3 x | + c

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\frac{1}{5} l o g | s e c 5 x | - \frac{1}{3} l o g | s e c 3 x | - \frac{1}{2} l o g | s e c x | + c

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Solution

The correct option is C $\frac{1}{5} l o g | s e c 5 x | - \frac{1}{2} l o g | s e c 2 x | - \frac{1}{3} l o g | s e c 3 x | + c$
$t a n 5 x = t a n (2 x + 3 x) = \frac{t a n 2 x + t a n 3 x}{1 - t a n 2 x t a n 3 x} \Rightarrow t a n 5 x t a n 3 x t a n 2 x = t a n 5 x - t a n 3 x - t a n 2 x ∴ \int t a n 5 x t a n 3 x t a n 2 x d x = \int t a n 5 x d x - \int t a n 3 x - d x - \int t a n 2 x d x = \frac{1}{5} l o g | s e c 5 x | - \frac{1}{3} l o g | s e c 3 x | - \frac{1}{2} l o g | s e c 2 x | + c$

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