Evaluating Definite Integral $I$

[juantheron]

Evaluation of $\displaystyle \int^{\frac{1}{2}}_{0}\frac{1}{(1-2x^2)\sqrt{1-x}}dx$

$\bf{Try::}$ Let $\displaystyle I = \int^{\frac{1}{2}}_{0}\frac{1}{(1-2x^2)\sqrt{1-x}}dx$ Put $1-x=t^2\;,$ Then $dx=-2tdt$

So $\displaystyle I = \int^{1}_{\frac{1}{2}}\frac{2t}{\left[1-2(1-t^2)^2\right]t}dt = -2\int^{1}_{\frac{1}{2}}\frac{1}{2t^4-4t^2+1}dt$

Now how can i proceed further, Help me

Thanks

 

[Opalg]

Evaluation of $\displaystyle \int^{\frac{1}{2}}_{0}\frac{1}{(1-2x^2)\sqrt{1-x}}dx$

$\bf{Try::}$ Let $\displaystyle I = \int^{\frac{1}{2}}_{0}\frac{1}{(1-2x^2)\sqrt{1-x}}dx$ Put $1-x=t^2\;,$ Then $dx=-2tdt$

So $\displaystyle I = \int^{1}_{\frac{1}{2}}\frac{2t}{\left[1-2(1-t^2)^2\right]t}dt = -2\int^{1}_{\frac{1}{2}}\frac{1}{2t^4-4t^2+1}dt$

Now how can i proceed further

This should work in principle, though I wouldn't want to try to do it by hand:
$$\begin{aligned}2t^4-4t^2+1 &= (2t^2-1)^2 -2t^4 \\ &= \bigl((2+\sqrt2)t^2 - 1\bigr)\bigl((2-\sqrt2)t^2 - 1\bigr) \\ &= \bigl(\sqrt{2 + \sqrt2}t + 1\bigr)\bigl(\sqrt{2 + \sqrt2}t - 1\bigr)\bigl(\sqrt{2 - \sqrt2}t + 1\bigr)\bigl(\sqrt{2 - \sqrt2}t - 1\bigr) \end{aligned}$$ Now use partial fractions!