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