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प्रश्न
Evaluate the following integral:
`int (3cosx)/(4sin^2x + 4sinx - 1).dx`
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उत्तर
Let I = `int (3cosx)/(4sin^2x + 4sinx - 1).dx`
Put sin x = t
∴ cosx dx = dt
∴ I = `int 3/(4t^2 + 4t - 1)dt`
I = `3/4 int 1/(t^2 + t - 1/4)dt`
I = `3/4 int 1/((t^2 + t + 1/4) - 1/4 - 1/4)dt`
I = `3/4 int 1/ ((t + 1/2)^2 - 1/2)dt`
I = `3/4 int 1/sqrt((t + 1/2)^2 - (1/sqrt2)^2)dt`
`[∵ int 1/(x^2 - a^2)dx = 1/(2a) log |(x - a)/(x + a)| + c]`
I = `3/4 xx 1/(2(1/sqrt2)) log |(t + 1/2 - 1/sqrt2)/(t + 1/2 + 1/sqrt2)| + c`
I = `3/(4sqrt2) log |(2sqrt2t + (2sqrt2)/2 - (2sqrt2)/sqrt2)/(2sqrt2t + (2sqrt2)/2 - (2sqrt2)/sqrt2)| + c`
I = `3/(4sqrt2) log |(2sqrt2t + sqrt2 - 2)/(2sqrt2t +sqrt2 + 2)| + c`
I = `3/(4sqrt2) log |(2sqrt2sin + sqrt2 - 2)/(2sqrt2sin +sqrt2 + 2)| + c`
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