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evaluate ∫sin−1x dx |
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Answer» We need to evaluate ∫ sin-1x dx Let u = sin-1x, dv = dx ⇒ du = \(\frac{1}{\sqrt{1-x^2}}\)dx, v = x ∫ sin-1x dx = x sin-1x - ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx + c ....(1) where c is integration constant. Consider ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx Let w = 1 - x2 ⇒ dw = -2xdx ⇒ xdx = -dw/2 ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx = ∫ \(-\frac{dw}{2\sqrt w}\) = -1/2 ∫ dw/√w = -1/2 (2√w) = -√w substitute back w = 1 - x2 we get ∴ ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx = \(-\sqrt{1-x^2}\) Therefore (1) becomes ∫ sin-1x dx = x sin-1 x + \(\sqrt{1-x^2}\) + c |
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