The given integral is
We have,
3sin x + 2cos x + 4
= l(3cos x + 4 sin x + 5) + m(– 3sin x + 4cos x) + n
= (4l – 3m)sin x + (3l + 4m) cos x + (5 l + n)
Comparing the co-efficients of sin x and cos x, we get
= 4l – 3m = 3, 3 l + 4m = 2 and 5l + n = 4
Solving, we get
l = 18/25, m = –1/25, n = 1/5
Thus,
