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Evaluation of $\displaystyle \int\frac{1}{2x\sqrt{1-x}\cdot \sqrt{2-x+\sqrt{1-x}}}dx$

$\bf{My\; Try:}$ Let $x=\sin^2 \theta\;,$ Then $dx = 2\sin \theta \cdot \cos \theta d\theta$

So Integral $$\displaystyle I = \int\frac{2\sin \theta \cos \theta }{2\sin^2 \theta \cos \theta\sqrt{2-\sin^2 \theta+\cos \theta}}d\theta = \int\frac{1}{\sin \theta\sqrt{\cos^2 \theta+\cos \theta+1}}d\theta$$

So $$\displaystyle I = \int\frac{\sin \theta}{(1-\cos^2 \theta)\sqrt{\cos^2 \theta+\cos \theta+1}}d\theta\;,$$ Now put $\cos \theta = t\;,$ Then $\sin \theta d\theta = -dt$. So $$\displaystyle I = \int\frac{1}{(1-t^2)\sqrt{t^2+t+1}}dt$$

How can we continue?

wythagoras
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juantheron
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2 Answers2

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I give you a step or two, and know you can finish from them:

Make a partial fraction decomposition of the rational part, and write your integral as $$ \frac{1}{2}\int\frac{1}{(1-t)\sqrt{1+t+t^2}}\,dt+\frac{1}{2}\int\frac{1}{(1+t)\sqrt{1+t+t^2}}\,dt $$ In the first integral, let $$ u=\frac{1-t}{\sqrt{1+t+t^2}}, $$ and in the second, let $$ u=\frac{1+t}{\sqrt{1+t+t^2}}. $$ You will end up with integrals of the form $$ \int\frac{1}{c-u^2}\,du $$ which I'm certain you can handle.

mickep
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Nine years too late!!

After decomposing $1-t^2$ in the denominator of the integrand into $1-t$ and $1+t$, perform the substitutions $1-t=\frac1{u}$ and $1+t=\frac1{u}$ respectively:

$$\int\frac{\mathrm dt}{(1-t^2)\sqrt{t^2+t+1}}=\frac12\int\frac{\mathrm dt}{(1-t)\sqrt{t^2+t+1}}+\frac12\int\frac{\mathrm dt}{(1+t)\sqrt{t^2+t+1}}$$

$$\int\frac{\mathrm dt}{(1-t)\sqrt{t^2+t+1}}=-\int\frac{\mathrm du}{\sqrt{3u^2-3u+1}}$$

$$\int\frac{\mathrm dt}{(1+t)\sqrt{t^2+t+1}}=-\int\frac{\mathrm du}{\sqrt{u^2-u+1}}$$

Integreek
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