# How do I integrate this?

How do I integrate $\displaystyle\int (x^2 + 2)\sqrt{1-x} \; dx$ ?

I have feeling substitution might be used, but I just can't put my finger on it...

Thank you.

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have you tried trig substitutions? – Holdsworth88 Mar 5 '12 at 10:28
substitute $u=\sqrt{1-x}$ so, $du=-\frac{1}{2\sqrt{1-x}}$ – Riccardo.Alestra Mar 5 '12 at 10:30
@Pjennings Nope, I don't really know how it would be applied in this case... – kralyk Mar 5 '12 at 10:31
@Riccardo.Alestra That's what wolfram says, but I don't get what to do next. Sorry, I possibly lack the skill :/ – kralyk Mar 5 '12 at 10:33

The ‘problem child’ in your integrand is $\sqrt{1-x}$; that should immediately suggest substituting either $u=\sqrt{1-x}$ or $u=1-x$. The latter looks simpler, so let’s try it and see what happens. We get $du=-dx$, which is nice and simple, and it’s easy enough to solve for $x$ to find that $x=1-u$. Now substitute $1-u$ for $x$ in the rest of the integrand, and you get

\begin{align*}\int (x^2+2)\sqrt{1-x}\,dx&=-\int\left((1-u)^2+2\right)\sqrt u\, du\\ &=-\int(3-2u+u^2)u^{1/2}\,du\;. \end{align*}

Now just multiply out, use the power rule, and reverse the substitution.

You might wonder what would have happened if we’d used the substitution $u=\sqrt{1-x}$ instead. Then we’d have $x=1-u^2$, so $x^2+2=(1-u^2)^2+2=3-2u^2+u^4$, which isn’t bad. We also get $$du=\frac{-1}{\sqrt{1-x}}dx\;,$$ or $dx=-\sqrt{1-x}\,du$ which looks a little ugly until you realize that it’s just $-u\,du$. Thus, with this substitution we get

$$\int (x^2+2)\sqrt{1-x}\,dx=-\int(3-2u^2+u^4)u^2\, du\;,$$

which turns out to be not so bad after all.

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Thanks! Finally got it. – kralyk Mar 5 '12 at 10:57

Note that $$x^2+2=(1-x)^2-2(1-x)+3,$$ which implies that $$(x^2+2)\sqrt{1-x}=(1-x)^\frac{5}{2}-2(1-x)^{\frac{3}{2}}+3(1-x)^{\frac{1}{2}}.$$ Therefore, let $u=1-x$, we have $dx=-du$, which implies that $$\int (x^2+2)\sqrt{1-x}dx=-\int u^\frac{5}{2}-2u^{\frac{3}{2}}+3u^{\frac{1}{2}}du$$ $$=-\frac{2}{7}u^{\frac{7}{2}}+\frac{4}{5}u^{\frac{5}{2}}+2u^{\frac{3}{2}}+C$$ $$=-\frac{2}{7}(1-x)^{\frac{7}{2}}+\frac{4}{5}(1-x)^{\frac{5}{2}}+2(1-x)^{\frac{3}{2}}+C.$$

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Beat me to it. +1 – user21436 Mar 5 '12 at 10:40
-1, This it too slick for such a simple integration problem – Norbert Mar 5 '12 at 10:41
(+1) Nice and clean solution – TMM Mar 5 '12 at 10:52
It works, but it’s not particularly helpful in the long run, because it gives no indication of where the first displayed line came from. – Brian M. Scott Mar 5 '12 at 10:53
+1 very nice!!! – draks ... Mar 5 '12 at 11:45

Another way is a rationalizing substitution ("rationalizing" $=$ getting rid of the radical): $$u=\sqrt{1-x}$$ $$u^2=1-x$$ $$2u\;du = -dx$$ $$x=1-u^2$$ $$\int (x^2 + 2)\sqrt{1-x} \; dx = \int((1-u^2)^2+2) u (2u\;du).$$ Then multiply it out and you're integrating a polynomial.

Finally, but $\sqrt{1-x}$ in place of each "$u$" at the end.

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