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प्रश्न
`int 1/(sin x - sqrt3 cos x) dx`
योग
उत्तर
Given I = `int 1/(sin x - sqrt3 cos x) dx`
Let 1 = r cos θ and √3 = r sin θ
r = `sqrt(3 + 1) = 2`
And tan θ = √3 → θ = `pi/3`
=> `int 1/(sin x - sqrt3 cos x) dx = int 1/(rcos theta sin x - r sin theta cos x) dx`
= `1/r int 1/(sin (x - theta))dx`
= `1/r int cosec(x - theta)dx`
We know that `int cosec x dx = log|tan (x/2 - pi/6)| + c`
`1/2 log |tan(x/2 - pi/6)| + c`
∴ I = `int 1/(sinx - sqrt3 cos x) dx`
`1/2 log |tan (x/2 - pi/6)| + c`
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