Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 9 down vote accepted

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.

share|cite|improve this answer
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 $$\int (x^2+2)\sqrt{1-x}dx=-\int u^\frac{5}{2}-2u^{\frac{3}{2}}+3u^{\frac{1}{2}}du$$ $$=-\frac{2}{7}u^{\frac{7}{2}}+\frac{4}{5}u^{\frac{5}{2}}+2u^{\frac{3}{2}}+C$$ $$=-\frac{2}{7}(1-x)^{\frac{7}{2}}+\frac{4}{5}(1-x)^{\frac{5}{2}}+2(1-x)^{\frac{3}{2}}+C.$$

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.