Question Number 10510 by ajfour last updated on 15/Feb/17

Answered by mrW1 last updated on 15/Feb/17

2sin x+3cos x  =(√(13))×((2/(√(13)))sin x+(3/(√(13)))cos x)  =(√(13))×(sin xcos t+cos xsin t)  =(√(13))×sin (x+t)  with t=sin^(−1) ((3/(√(13)))) and tan t=(3/2)    ∫(1/((2sin x+3cos x)^2 ))dx  =∫(1/(13sin^2  (x+t)))dx  =(1/(13))×∫(1/(sin^2  (x+t)))d(x+t)  =(1/(13))×[−cot (x+t)]+C  =(1/(13))×(−((1−tan x tan t)/(tan x+tan t)))+C  =(1/(13))×((−1+(3/2)tan x)/(tan x+(3/2)))+C  =(1/(13))×((3tan x−2)/(2tan x+3))+C  =(1/(13))×((3sin x−2cos x)/(2sin x+3cos x))+C

Answered by robocop last updated on 15/Feb/17

=ln∣(2sinx+3cosx)^2 ∣