11Can anyone do this one .....∫ (tanθ)3/2dθ
7 Answers
1dude its a standard integral of the form x^2 - a^2 where a is constant
in your case a=1 and x^16=x^2^8
39monkey man,
i am really waiting to see how you would do the way u said. can u post full solution.
19monkey....really the name suggests that.....look at (dx)....for the way you are suggesting it should have been d(x8) because x16 = ( x8)2
11@sum
write it as ∫tan@√tan@d@
put tan@=t and sec^2 @ d@=dt and also sec^2 @ =1+tan^2 @
use both and get ur answer
and vimalesh
u need ∫1x^16-1dx
so put x=sec1/8@ dx=1/8 sec-7/8d@
ab kya reh gaya iss mein
solve it
time lagega :|