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When i integrate $\int_{0}^{1}x^{7}\sqrt{\frac{1+x^{2}}{1-x^{2}}}dx$, putting $t^{2}=1-x^{2}$ then integration $\int{(1-t^{2})}^{3}\sqrt{t^{2}+2}dt$, but i don't understand how to proceed . please someone help me. Thank you.

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    the solutions looks ugly. – 2017-01-14
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    Can you start with $x^2=\cos2t$ – 2017-01-14
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    the solution is not that ugly, it equals $\displaystyle\frac{9 \pi +32}{96}$ – 2017-01-14

3 Answers 3

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One way to proceed further is an Euler substitution,

$$u=\sqrt{t^2+2}-t$$ this will reduce the integral to a rational function.

Another way to proceed from the beginning is to set $y=x^2$ then you get

$$\frac{1}{2}\int y^3\sqrt{\frac{1+y}{1-y}}dy$$

and now the substitution $$z=\sqrt{\frac{1+y}{1-y}}$$ will again give a rational expression, this method is probably neater.

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    Thank you. But what will be limit – 2017-01-14
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Substitute $\text{u}=\frac{1}{1-x^2}$:

$$\mathcal{I}=\int x^7\cdot\sqrt{\frac{1+x^2}{1-x^2}}\space\text{d}x=\frac{1}{2}\int\frac{\left(\text{u}-1\right)^3\cdot\sqrt{2\text{u}-1}}{\text{u}^5}\space\text{d}\text{u}$$

Now, substitute $\text{v}=\sqrt{2\text{u}-1}$:

$$\mathcal{I}=2\int\frac{\text{v}^2\cdot\left(\text{v}^6-3\text{v}^4+3\text{v}^2-1\right)}{\left(1+\text{v}^2\right)^5}\space\text{d}\text{v}$$

And now you do partial fraction decomposition:

$$\frac{\text{v}^2\cdot\left(\text{v}^6-3\text{v}^4+3\text{v}^2-1\right)}{\left(1+\text{v}^2\right)^5}=\frac{1}{1+\text{v}^2}-\frac{7}{\left(1+\text{v}^2\right)^2}+\frac{18}{\left(1+\text{v}^2\right)^3}-\frac{20}{\left(1+\text{v}^2\right)^4}+\frac{8}{\left(1+\text{v}^2\right)^5}$$

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    Nice, by the way that is the same result as my second suggestion, I was too lazy to do the partial fraction expansion, so thanks. But integrating those fractions is not easy. – 2017-01-14
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    @ReneSchipperus Haha, you're welcome. I was writing the same answer as you, so when I saw your answer I changed it :) (Btw. are you Dutch? Your name sounds Dutch to me xD) – 2017-01-14
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    Well spotted, my father was Dutch. It means one who ships, names were given after professions back then. Are you also Dutch ? – 2017-01-14
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    @ReneSchipperus Ahh oke, yes I'm Dutch born and raised :). I think that your name is not well spoken abroad (here in Holland we can say 'sch' sounding as a 'g') – 2017-01-14
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    Interestingly, the Germans have a lot of trouble with the name. – 2017-01-14
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    @ReneSchipperus Haha, oke that's new to me :) where do you live now (if I may ask)? – 2017-01-14
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    If I knew the answer to that I would give it. – 2017-01-14
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HINT:

For $$\int x^{2n+1}f(x^2)dx$$ I'll start with $x^2=u$

let $\sqrt{\dfrac{1+x^2}{1-x^2}}=u\implies x^2=\dfrac{u^2-1}{u^2+1}=1-\dfrac2{1+u^2}$