Question

̣ tan 2x tan 3x tan 5x dx is equal to –

• A. $\frac{1}{2}$log|sec 2x| – $\frac{1}{3}$log|sec 3x| – $\frac{1}{5}$log|sec 5x|+C
• B. $\frac{1}{2}$log|sec 2x|+$\frac{1}{3}$log|sec 3x|+$\frac{1}{5}$log|sec 5x|+C
• C. $\frac{1}{5}$log|sec5x|–$\frac{1}{2}$log|sec 2x|–$\frac{1}{3}$log|sec 3x|+C
• D. None of the above

Answer 1

Answer: (C)

$\frac{1}{5}$log|sec5x|–$\frac{1}{2}$log|sec 2x|–$\frac{1}{3}$log|sec 3x|+C

Answer 2

tan 5x = tan (3x + 2x) = $\frac{\tan 2x + \tan 3x}{1–\tan 2x.\tan 3x}$

̃ tan 5x – tan2x. tan 3x. tan 5x = tan 2x + tan 3x

̃ ̣ tan2x. tan3x. tan5x dx = ̣ tan5x – tan2x – tan3x. dx

Now solve.........