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How do I find $\int \sqrt{(26x-x^2)} dx $

Is this an integration by parts question?

Thanks,

--Nick

user2588096
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  • I corrected the latex formatting. Can the OP please check that the formula is correct? Is the square root over the entire argument? Ok... I see editing is ongoing :-) – Umberto Apr 18 '14 at 07:59
  • Sorry I may have just undone your work. The square root is around 26x-x^2. I am new to this. – user2588096 Apr 18 '14 at 08:01
  • No problem. You can write the square root in latex in this way: \sqrt{} with curly braces. Everything in the curly braces will be below the square root. – Umberto Apr 18 '14 at 08:01
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    This may be helpful http://math.stackexchange.com/questions/390080/definite-integral-of-square-root-of-polynomial – Umberto Apr 18 '14 at 08:04

4 Answers4

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Hint: First notice that $26x-x^{2}=-(x^{2}-26x)=-((x-13)^{2}-169)=169-(x-13)^{2}$

user71352
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Integrate as follows

\begin{align} \int \sqrt{26x-x^2} dx &= \int \frac{\sqrt{26x-x^2}}{2(x-13)}d[(x-13)^2]\\ &=\frac12(x-13)\sqrt{26x-x^2}+ \frac{169}2\int\frac1{ \sqrt{26x-x^2}}dx\\ &= \frac12(x-13)\sqrt{26x-x^2}+\frac{169}2\sin^{-1}{\frac {x-13}{13}}+C\\ \end{align}

Quanto
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Using Trigonometric Substitution, $$2ax-x^2=a^2-(x-a)^2$$

Set $$x-a=a\sin\theta$$

Here $a=13$

lab bhattacharjee
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Note \begin{align}\int{\sqrt{26x-x^2}dx}& = \int{\sqrt{169-(x^2-26x+169)}dx} \\ &=\int{\sqrt{13^2-(x-13)^2}dx} \end{align} It can be evaluated using $x-13=13\sin t,0\leq t \leq \pi/2$.

Ángel Mario Gallegos
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