Aditya Gupta
Last Activity: 6 Years ago
put tanx=y
we get dy=(1+y^2)dx
or dy/[(1+y^2)(1+ay)]
now simply write in form of partial fractions:
1/[(1+y^2)(1+ay)]= c/(1+ay) + (fy+e)/(1+y^2) and determine the coefficients c d and f
c/(1+ay) is integrated as cln(1+ay)/a
write (fy+e)/(1+y^2) as (fy)/(1+y^2) + (e)/(1+y^2)
(e)/(1+y^2) has integral e*arctany.
(fy)/(1+y^2) can be solved by letting 1+y^2=z so its integral is
f/2*ln(1+y^2)
add all these together now and thats it