Ngô Quốc Anh

June 5, 2008

Nhiều lời giải cho một bài tích phân xác định chứa căn thức…

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 4:58

Evaluate

\displaystyle I = \int_0^1 {x\sqrt {\frac{{1 - x}}{{1 + x}}} dx}.

Solution 1. Put x=\cos(2y) to get

\displaystyle\begin{gathered} I = \int_0^{\pi /4} {2\cos 2y\frac{{\sin y}}{{\cos y}}\ \sin 2y\ dy} \ \hfill \\ \quad= \int_0^{\pi /4} {4{{\sin }^2} ydy\ } - \frac{1}{4}\int_0^{\pi /4} {\ 32{{\sin }^4} ydy} \hfill \\ \quad= \left( {2y - \sin 2y \bigg |_0^{\pi /4}} \right) - \frac{1}{4}\left( {12y - 8\sin 2y + \sin 4y\bigg |_0^{\pi /4}} \right) \hfill \\ \quad= 1 - \frac{\pi }{4}. \hfill \\ \end{gathered}

Solution 2. We do the following calculation

\displaystyle\begin{gathered} I = \int_0^1 {\frac{x}{{x + 1}}\sqrt {1 - {x^2}} dx} \hfill \\ \quad= \int_0^1 {\frac{{(x + 1) - 1}}{{x + 1}}\sqrt {1 - {x^2}} dx} \hfill \\ \quad= \int_0^1 {\sqrt {1 - {x^2}} dx} - \int_0^1 {\frac{{\sqrt {1 - {x^2}} }}{{x + 1}}dx} \hfill \\ \quad= \frac{\pi }{4} - \int_0^1 {\sqrt {\frac{{1 - x}}{{1 + x}}} dx} \hfill \\ \quad= \frac{\pi }{4} - \int_0^1 {\frac{{1 - x}}{{\sqrt {1 - {x^2}} }}dx} \hfill \\ \quad= \frac{\pi }{4} - \int_0^1 {\frac{1}{{\sqrt {1 - {x^2}} }}dx} + \int_0^1 {\frac{x}{{\sqrt {1 - {x^2}} }}dx} {\text{ d}}x \hfill \\ \quad= - \arcsin x - \sqrt {1 - {x^2}} \bigg|_0^1 + \frac{\pi }{4} \hfill \\ \quad= 1 - \frac{\pi }{4}. \hfill \\ \end{gathered}

Solution 3. Since

\displaystyle 0 \leqslant \theta \leqslant \frac{\pi }{2}

we have by x = \sin \theta that

\displaystyle\begin{gathered} \int_0^1 {\frac{1}{{\sqrt {1 - {x^2}} }}} dx = \int_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {1 - {{\sin }^2}\theta } }}\cos \theta d\theta } \hfill \\ \qquad= \int_0^{\frac{\pi }{2}} {\frac{1}{{|\cos \theta |}}\cos \theta d\theta } \hfill \\ \qquad= \int_0^{\frac{\pi }{2}} {\frac{1}{{\cos \theta }}\cos \theta d\theta } \hfill \\ \qquad= \frac{\pi }{2}. \hfill \\ \end{gathered}

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