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