Evaluate ∫√((1-x)/x) dx\(\int\sqrt{\frac{1-x}{x}}\) dx

1.	Evaluate ∫√((1-x)/x) dx\(\int\sqrt{\frac{1-x}{x}}\) dx
Answer» Given, ∫√((1-x)/x) dx Let, \(\sqrt x\) = t \(\frac{d}{dx}(\sqrt x)\) = dt \(\frac{1}{2\sqrt x}\) dx = dt dx = 2t dt Now, \(\int\frac{\sqrt {1-t^2}}{t}\) 2t dt = 2\(\int\sqrt{1-t^2}\) dt Consider, t = sin k dt = cos k dk = 2\(\int\sqrt{1-sin^2k}\) .cosk dk = 2\(\int\sqrt{cos^2k}\) .cosk dk = 2 ∫cos²k dk = ∫2 cos²k dk =∫cos 2k-1 dk [since, cos 2x = 2cos²x-1] = \(\frac{sin\,2k}{2}\) - k + c = \(\frac{2sink\,cosk}{2}\) - k + c = t cos(sin^-1t) - 2sin^-1t + 2c =√x cos(sin^-1√x )- 2sin^-1√x + 2c

Answer»

Given,

∫√((1-x)/x) dx

Let,

\(\sqrt x\) = t

\(\frac{d}{dx}(\sqrt x)\) = dt

\(\frac{1}{2\sqrt x}\) dx = dt

dx = 2t dt

Now,

\(\int\frac{\sqrt {1-t^2}}{t}\) 2t dt

= 2\(\int\sqrt{1-t^2}\) dt

Consider,

t = sin k

dt = cos k dk

= 2\(\int\sqrt{1-sin^2k}\) .cosk dk

= 2\(\int\sqrt{cos^2k}\) .cosk dk

= 2 ∫cos²k dk

= ∫2 cos²k dk

=∫cos 2k-1 dk

[since, cos 2x = 2cos²x-1]

= \(\frac{sin\,2k}{2}\) - k + c

= \(\frac{2sink\,cosk}{2}\) - k + c

= t cos(sin^-1t) - 2sin^-1t + 2c

=√x cos(sin^-1√x )- 2sin^-1√x + 2c

Discussion