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प्रश्न
`int (sinx)/(sin3x) "d"x`
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उत्तर
Let I = `int (sinx)/(sin3x) "d"x`
= `int sin x/(3sin x - 4 sin^3x)* "d"x`
= `int sinx/(sinx(3 - 4sin^2x))* "d"x`
= `int 1/(3 - 4sin^2x) "d"x`
Dividing numerator and denominator by cos2x, we get
I = `int (sec^2x)/(3sec^2x - 4tan^2x) * "d"x`
= `int (sec^2x)/(3(1 + tan^2x) - 4tan^2x)* "d"x`
= `int (sec^2x)/(3 - tan^2x) "d"x`
Put tan x = t
∴ sec2x dx = dt
∴ I = `int "dt"/(3 - "t"^2)`
= `int 1/((sqrt(3))^2 - "t"^2) "dt"`
=`1/(2sqrt(3)) log|(sqrt(3) + "t")/(sqrt(3) - "t")| + "c"`
∴ I = `1/(2sqrt(3)) log|(sqrt(3) + tanx)/(sqrt(3) - tanx)| + "c'`
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