HP 15C and INT(1/√(1-x),0,1)
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07-20-2018, 01:04 PM
Post: #55
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RE: HP 15C and INT(1/√(1-x),0,1)
The HP-15C ADVANCED FUNCTIONS HANDBOOK has on page 47 a section about:
Transformation of Variables Quote:In many problems where the function changes very slowly over most of a very wide interval of integration, a suitable transformation of variables may decrease the time required to calculate the integral. You can use \(x=1-t^2\) which leads to: \[ \begin{eqnarray} dx&=&-2t\cdot dt \\ \sqrt{1-x}&=&t \\ \int_0^1\frac{1}{\sqrt{1-x}}dx&=&\int_1^0\frac{-2t}{t}dt \\ &=&2\int_0^1dt \\ &=&2t\bigg\vert_0^1 \\ &=&2-0=2 \end{eqnarray} \] Or then use \(x=\cos^2(t)\) which leads to: \[ \begin{eqnarray} \\ dx&=&-2\cos(t)\sin(t)dt \\ \sqrt{1-x}&=&\sin(t) \\ \int_0^1\frac{1}{\sqrt{1-x}}dx&=&\int_{\frac{\pi}{2}}^0\frac{-2\cos(t)\sin(t)}{\sin(t)}dt \\ &=&2\int_0^{\frac{\pi}{2}}\cos(t)dt \\ &=&2\sin(t)\bigg\vert_0^{\frac{\pi}{2}} \\ &=&2-0=2 \end{eqnarray} \] In both cases the resulting function can be integrated without problems. It's just a coincidence that both integrals can also be calculated easily algebraically. Cheers Thomas |
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