Question

Integrate the following functions w.r.t. x : `(sinx + 2cosx)/(3sinx + 4cosx)`

Answer 1

Let I = `int (sinx + 2cosx)/(3sinx + 4cosx).dx`
Put,
Numberator = `"A (Denominator) + B"[d/dx("Denominator")]`

∴ sinx + 2cosx = `"A"(3sinx + 4cosx) + "B"[d/dx(3sinx + 4cosx)]`

= A(3 sin x + 4 cos x) + B(3 cos x – 4 sin x)
∴ sin x + 2 cos x = (3A –  4B)sin x + (4A + 3B)cos x
Equaliting the coefficients of sin x and cos x on both the sides, we get
3A – 4B = 1     ...(1)
and
4A + 3B = 2   ...(2)
Multiplying equation (1) by 3 and equation (2) by 4, we get
9A – 12B = 3
16A + 12B = 8
On adding, we get
25A = 11

∴ A = `(11)/(25)`

∴ from (2), `4(11/25) + 3"B"` = 2

∴ 3B = `2 - (44)/(25) = (6)/(25)`

∴ B = `(2)/(25)`

∴ `sinx + 2cos x = (11)/(25)(3sinx + 4cosx) + (2)/(25)(3cosx - 4sinx)`

∴ I = `int[(11/25(3sinx + 4cosx) + 2/25(3cosx - 4sinx))/(3sinx + 4cosx)].dx`

= `int[11/25 + (2/25(3cosx - 4sinx))/((3sinx + 4cosx))].dx`

= `(11)/(25) int1 dx + 2/25 int(3cosx - 4sin x)/(3sin x + 4cosx).dx`

= `(11)/(25)x + (2)/(25)log|3 sin x + 4 cos x| + c.     ...[∵ int (f'(x))/(f'(x))dx = log|f(x)|  +  c]`