If the substitution square root of $x = \sin y$ is made in the $\int^0_{0.5}\dfrac{\sqrt{x}}{\sqrt{1-x}}dx$, what is the resulting integral?
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Do you mean $$\int_{.5}^0\sqrt{\frac{x}{1-x}}dx$$? – Cameron Williams Mar 10 '14 at 02:07
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@CameronWilliams It's the same thing. – Mar 10 '14 at 02:10
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Are you sure the lower limit is $0.5$? (usually one uses the smaller number as lower limit, and if not, switch them and change sign) – coffeemath Mar 10 '14 at 02:51
1 Answers
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I will do $0$ to $\frac{1}{2}$, i.e., $\displaystyle \int_0^{0.5}\frac{\sqrt{x}}{\sqrt{1-x}}dx$.
First, as $g(x)=\frac{\sqrt{x}}{\sqrt{1-x}}$ is continuous in $(0,\frac{1}{2})$, it follows that this integral is not unfit.
Now, replace $1-x =u^2$, then $x=1-u^2$ and $dx=-2udu$.
Calculating the indefinite integral $$\int \frac{\sqrt{x}}{\sqrt{1-x}}dx=-2\int \frac{\sqrt{1-u^2}\cdot u}{u}du =-2\int \sqrt{1-u^2} du =(*).$$
Now, do $$\sin t=\sqrt{1-u^2} \text{ and } \cos t=u,$$ then $du=-\sin t dt$ and so $$(*)=-2\int \sin t(-\sin t)dt=2\int \sin^2t dt=2\int \frac{1-\cos 2t}{2}dt= $$ $$=\int dt -\int \cos 2t dt=t-\frac{1}{2}\sin 2t+c=t-\sin t\cos t +c= $$ $$=\arccos u-u\sqrt{1-u^2}+c=\arccos\sqrt{1-x}-\sqrt{1-x}\cdot\sqrt{x}+c. $$
Lastly,
$$\int_0^{\frac{1}{2}}g(x)dx= \left.(\arccos\sqrt{1-x}-\sqrt{1-x}\cdot\sqrt{x})\right|_0^{\frac{1}{2}}= $$ $$=\arccos \frac{\sqrt{2}}{2}-\sqrt{\frac{1}{2}}\cdot \sqrt{\frac{1}{2}}-\left( \arccos 1-0\right) =\frac{\pi-2}{4}.$$
ZHN
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