Question

Integrate the following functions w.r.t. x : `(1)/(2 + 3tanx)`

Answer 1

Let I = `int (1)/(2 + 3tanx).dx`

= `int(1)/(2 + 3(sinx/cosx)).dx`

= `int cosx/(2cosx + 3sinx).dx`
Put,

Numerator = `"A (Denominator) + B"[d/dx("Denominator")]`

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

= A(2 cos x + 3 sin x) + B(– 2 sin x + 3 cos x)

∴ cos x = (2A + 3B)cos x + (3A – 2B)sin x
Equating the coefficients of cos x sin x on both the sides, we get
2A  3B = 1          ...(1)
and
3A – 2B = 0      ...(2)
Multiplying equation (1) by 22 and equation (2) by 3, we get
4A +6B = 2
9A – 6B = 0
On adding, we get
13A = 2
∴ A = `(2)/(13)`

∴ from (2), 2B = 3A = `3(2/13) = (6)/(13)`

∴ B = `(3)/(13)`

∴ cos x = `(2)/(13)(2cosx + 3sinx) + (3)/(13)(-2sinx +  3cosx)`

∴ I = `int[(2/13(2cosx + 3sinx) + 3/13(-2 sinx + 3cosx))/(2cosx + 3sinx)].dx`

= `int[2/13 + (3/13(-2sinx + 3cosx))/(2cosx + 3sinx)].dx`

= `(2)/(13) 1 dx + (3)/(13) int (-2sinx + 3cosx)/(2cosx + 3sinx).dx`

= `(2)/(13)x + (3)/(13)log|2cos x + 3sinx| + c.     ...[∵ int (f'(x))/f(x)dx = log|f(x)| + c]`