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For example, I have the problem:

$$ \int_{-1}^1x^4\left(1-x^2\right)^2dx $$ I set $\,u=1-x^2\,$, which gives $\,x^2=1-u\,$, $\,x=\pm\sqrt{1-u}\,$ and $\,-\frac{1}{2}du=xdx$.

For the upper and lower bounds of the integral:

$$ x=-1 \rightarrow u=1-\left(-1\right)^2=1-1=0 $$

$$ x=1 \rightarrow u=1-\left(1\right)^2=1-1=0 $$

By using u-substitution, I arrive at

$$ -\frac{1}{2}\int_0^0\pm\sqrt{1-u}\left(1-u\right)u^2du $$

We know that $$ \int_a^af(x)dx = 0 $$

I came to the conclusion that this integral is equal to zero as the upper and lower bounds are equal, however I know that that is not true.

I found other questions which address this subject, but none seem to make since to me as to why this is not true.

I am aware that u-substitution is unnecessary for this solution, but I am unclear as to why it doesn't seem to work for $u=1-x^2$. I suspect that the issue arises in the square root function, but I'm not sure, and would like to know how to avoid incorrect solutions like this in the future.

Thank-you.

Liam
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    The range of $u$s should match the range of $x$s. You need an interval of $u$s such that each $x\in[-1,1]$ matches with one and only one $u$. Such an interval does not exist in your case. However, you can observe that the original integrand is even. Hence $$\int_{-1}^1x^4(1-x^2)^2,dx=2\int_0^1x^4(1-x^2)^2,dx.$$ You can use your substitution for the integral on the right hand side as $x\in[0,1]$ corresponds to $u\in[0,1]$ (but in the opposite direction). It doesn't look like this substitution helps in evaluating the integral, but that was not the point. – Jyrki Lahtonen Oct 09 '18 at 04:25
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    Lots of similar questions have been asked before: https://math.stackexchange.com/questions/829939/why-does-this-u-substitution-zero-out-my-integral, https://math.stackexchange.com/questions/2380669/proof-all-integrals-are-0, https://math.stackexchange.com/questions/37446/when-a-change-of-variable-results-in-equal-limits-of-integration, https://math.stackexchange.com/questions/1088287/change-of-variable-in-integration-causes-upper-and-lower-unequal-limits-to-be-sa, etc. – Hans Lundmark Oct 09 '18 at 06:36

3 Answers3

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To use the substitution $u=1-x^2$ to evaluate, $$ \int_{-1}^1x^4\left(1-x^2\right)^2\,\mathrm{d}x $$ one has to remember that for $x\in[0,1]$, we have $x=\sqrt{1-u}$ giving $\mathrm{d}x=\frac{-\mathrm{d}u}{2\sqrt{1-u}}$ and for $x\in[-1,0]$, we have $x=-\sqrt{1-u}$ giving $\mathrm{d}x=\frac{\mathrm{d}u}{2\sqrt{1-u}}$. Thus, we get $$ \int_{-1}^0x^4\left(1-x^2\right)^2\,\mathrm{d}x=\int_0^1(1-u)^2u^2\frac{\mathrm{d}u}{2\sqrt{1-u}} $$ and $$ \begin{align} \int_0^1x^4\left(1-x^2\right)^2\,\mathrm{d}x &=\int_{\color{#C00}{1}}^{\color{#C00}{0}}(1-u)^2u^2\frac{\color{#C00}{-}\mathrm{d}u}{2\sqrt{1-u}}\\ &=\int_0^1(1-u)^2u^2\frac{\mathrm{d}u}{2\sqrt{1-u}} \end{align} $$

robjohn
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As said by @JyrkiLahtonen in comment, you have to use the substitution which corresponds to the range of the variable of your integration.

The thing is that's why you're getting the value of integral 0, because $u$ has range from [0,1] and x has range from [-1,1].

This integration, if you don't want to make some substitution can be done by simple solving the question, by squaring and multiplying the contents.

You'll get 3 term with $x^n$ type and that will be easy to solve. I think the answer to this integral without limits will be

$$\frac {x^9} 9 - \frac {2x^7} 7 + \frac {x^5} 5$$

PradyumanDixit
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$$I=\int_{-1}^1x^4(1-x^2)^2dx$$ notice that if we let $$f(x)=x^4(1-x^2)^2$$ then $$f(-x)=f(x)$$ and so the integral is equal to $$2\int_0^1x^4(1-x^2)^2dx$$

Henry Lee
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