so dy/dx = cos4x, therefore the integral of 41cos4x = 161sin4x
my final answer = 41x−161sin4x+c
but the answer book says its supposed to be: 81x−321sin4x+c
can someone tell me where i've gone wrong please.... ------------------------------------------------------------------ I've also got this to integrate: (secxtanx)^2 But I have no idea how to do it.........
so dy/dx = cos4x, therefore the integral of 41cos4x = 161sin4x
my final answer = 41x−161sin4x+c
but the answer book says its supposed to be: 81x−321sin4x+c
can someone tell me where i've gone wrong please.... ------------------------------------------------------------------ I've also got this to integrate: (secxtanx)^2 But I have no idea how to do it.........
Nothing major - you just missed a factor of 1/2:
(21sin2x)2=41sin2(2x)
As for (sec x tan x)^2, try using the substitution
but just out of interest, if I did you substitution as my method to integrate (secxtanx)^2 as others have suggested, would I still have got to the answer of '1/3tan^3x + c' ?
but just out of interest, if I did you substitution as my method to integrate (secxtanx)^2 as others have suggested, would I still have got to the answer of '1/3tan^3x + c' ?
i dont think you substitute otherwise you get to the power of 4 which is difficult to integrate...this is the method i know