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प्रश्न
Find: `int sin^-1 (2x) dx.`
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उत्तर
`int sin^-1 (2x) dx`
Using ILATE rule
`x sin^-1 2x - int (2x)/sqrt(1-4x^2)dx`
`x sin^-1 2x + 1/4 int (-8x)/sqrt(1-4x^2) dx`
Taking 1 - 4x2 = t
⇒ - 8x dx = dt
`x sin^-1 2x + 1/4 int dt/sqrt(t)`
`x sin^-1 2x + 1/4 (2t 1/2)/1 + "C"`
= `x sin^-1 2x + (t 1/2)/2 + "C"`
= `x sin^-1 2x + sqrt(1-4x^2)/2 + "C"`
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