3 Answers
Lokesh Verma
·2010-06-26 01:57:52
\int \frac{1}{1+4x^2}dx=1/2tan^{-1}(x/2)
So the given integral you have given is 1/4(tan-1(k/2) = pi/16
so tan-1(k/2) = pi/4
so ... k/2=1 so k=2
\int \frac{1}{1+4x^2}dx=1/2tan^{-1}(x/2)
So the given integral you have given is 1/4(tan-1(k/2) = pi/16
so tan-1(k/2) = pi/4
so ... k/2=1 so k=2