Ngô Quốc Anh

January 4, 2008

BT tính nguyên hàm

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 1:54

\int \frac{1}{\sin x \sqrt{\sin^2 x + a^2}} dx.

Solution. \int \dfrac { \sin \, x \; dx } { \sin^2 \, x \cdot \sqrt { a^2 + \sin^2 \, x } } \;=\;  \int \dfrac { \sin \, x \; dx } { ( 1 - \cos^2 \, x ) \cdot \sqrt { (a^2 + 1) - \cos^2 \, x } }

=\; \int \dfrac { \sec^2 \, x \;\tan\, x \; dx } { ( \sec^2 \, x - 1  ) \cdot \sqrt { (a^2 + 1 ) \sec^2 \, x  - 1} }.

Now put u^2 \;=\;(a^2 + 1 ) \sec^2 \, x  - 1

I\; =\;\dfrac { 1 } { a^2 + 1} \; \int  \dfrac { u\; du } { ( \dfrac { u^2 + 1} { a^2 + 1}  - 1 ) \cdot u } \;=\; \int \dfrac { du } { u^2-a^2}

 \;=\; \dfrac {1} {2a} \ln  \bigg| \dfrac {u-a} {u+a } \bigg | \qquad \text{or} \qquad -  \frac {1}{a}   \text{coth}^ {-1} \; \dfrac { u} { a}

\;=\;  \dfrac {1} {2a} \ln  \bigg| \dfrac { \sqrt {  a^2 + \sin^2 \, x  } -a\cos\, x } {\sqrt {  a^2 + \sin^2 \, x  } +a\cos\, x } \bigg | \qquad \text{or}        - \frac {1}{a}  \text{coth}^ {-1} \left(  \dfrac { \sqrt {   a^2 + \sin^2 \, x } } { a \cos\, x } \right).

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