inverse trig...

tan ^-1 (12r²)
can ne 1 tell hw do i break the above exp in a factor of 2 i.e., in the form of (a - b)

so dat i can evaluate the limits easily.....

2 Answers

\frac{(2r+1)-(2r-1)}{1+(2r-1)(2r+1)}=\frac{2}{4r^2}=\frac{1}{2r^2}

So tan^{-1}(\frac{1}{2r^2})=tan^{-1}(2r+1)-tan^{-1}(2r-1)

thanks nishant bhaiyya....
ur concepts r very strong....thank u very much

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