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Find: \(\int sin^{-1}2xdx\) |
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Answer» \(\int sin^{-1}2xdx\) = sin-12x \(\int 1dx\) - \(\int(\frac d{dx}sin^{-1}2x\int 1dx)dx\) = x sin-12x - \(\int\frac{2x}{\sqrt{1-4x^2}}dx\) = x sin-12x + \(\frac14\)\(\int\frac{-8x}{\sqrt{1-4x^2}}dx\) = x sin-12x + \(\frac14\)\(\int\frac{2tdt}t\) (By taking 1 - 4x2 = t2 ⇒ -8x dx = 2tdt) = x sin-12x + \(\frac12\sqrt{1-4x^2}+c\) (\(\because t=\sqrt{1-4x^2}\)) \(\therefore\) \(\int sin^{-1}2xdx\) = xsin-12x + \(\frac{\sqrt{1-4x^2}}2+c\) |
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