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∫dx2+3tanx - Mathematics and Statistics

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Question

`int ("d"x)/(2 + 3tanx)`

Sum
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Solution

Let I = `int 1/(2 + 3tanx)  "d"x`

= `int 1/(2 + 3(sinx/cosx))  "d"x`

= `int  cosx/(2cosx + 3sinx)  "d"x`

Let cos x = `"A"(2cosx + 3 sinx) + "B""d"/("d"x) (2cosx + 3sinx)`

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

∴ cos x + 0⋅sinx = cosx (2A + 3B) + sinx (3A − 2B)

By equating the coefficients on both sides, we get

2A + 3B = 1 and 3A − 2B = 0

Solving these equations, we get

A = `2/13` and B = `3/13`

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

∴ I = `int (2/13(2cos x + 3sin x) + 3/13(-2 sinx + 3cos x))/(2cosx + 3sin x)  "d"x`

∴ I = `2/13 int "d"x + 3/13 int (-2sinx + 3cosx)/(2cosx + 3sinx)  "d"x`

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

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Chapter 2.3: Indefinite Integration - Long Answers III

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