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Question

If the reduction formula for In=tannxdx is given by
In=1n1tann1xIn2, then tan3x dx is

A

12tan2x12ln(1+tan2x+C
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B

12tan2x12ln(1+tan2x+C
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C

12tan2x+12ln(1+tan2x+C
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D

12tan2x+12ln(1+tan2x+C
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Solution

The correct option is A
12tan2x12ln(1+tan2x+C
We know the reduction formulae for In=tannx dx as:
In=1n1tann1xIn2
Now, to find tan3x dx
we assume n=3.
Then, we can write the reduction formulae as:
I3=131tan31xI32
I3=12tan2xtanx dx
Now, we know
tan x dx = ln(secx|+Ctanx dx=ln(1+tan2x+CI1=12ln(1+tan2x+C
Thus, substituting for I1 we get,

tan3x dx=12tan2x 12ln(1+tan2x+C
Thus, The option a. is correct.

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