مثالهای حل انتگرال های تغییر متغیر (جانشانی) به کمک توابع مثلثاتی
انتگرالهای تغغیر متغیر به کمک توابع مثلثاتی بخش 1
1-[math]\int \frac{1}{x^{2}\sqrt{x^{2}-9}}dx[/math]
[math]x=3\sec \theta \rightarrow dx=3sec \theta\tan \theta \\
\sqrt{x^{2}-9}=\sqrt{9\sec ^2\theta -9}=\sqrt{9(\sec ^2\theta -1)}=\sqrt{9\tan ^2\theta }\\=3\tan \theta[/math]
[math] \int {\frac{1}{{{x^2}\sqrt {{x^2} – 9} }}} dx = \int {\frac{1}{{9{{\sec }^2}\theta .3\tan \theta }}} 3\sec \theta \tan \theta d\theta = \frac{1}{9}\int {\cos \theta d\theta = \frac{1}{9}} \sin \theta + C \\ = \frac{1}{9}\frac{{\sqrt {{x^2} – 9} }}{x} + C \\[/math]
سوال 2:
2-[math]\int \frac{x^{3}}{\sqrt{x^{2}+9}}dx[/math]
[math]x = 3\tan \theta \to dx = 3{\sec ^2}\theta d\theta \\ \sqrt {{x^2} + 9} = \sqrt {9{{\tan }^2}\theta + 9} = \sqrt {9({{\tan }^2}\theta + 1)} = \sqrt {9{{\sec }^2}\theta } = 3\sec \theta \\[/math]
[math]\int {\frac{{{x^3}}}{{\sqrt {{x^2} + 9} }}} dx = \int {\frac{{{3^3}{{\tan }^3}\theta }}{{3\sec \theta }}} 3e{c^2}\theta d\theta = {3^3}\int {{{\tan }^3}} \theta \sec \theta d\theta = {3^3}\int {{{\tan }^2}} \theta \tan \theta \sec \theta d\theta \\ = {3^3}\int {({{\sec }^2}} – 1)\tan \theta \sec \theta d\theta \\ \\{3^3}\int {({u^2}} – 1)du \leftrightarrow \left[ {u = \sec \theta ,du = \sec \theta \tan \theta d\theta } \right] \\ {3^3}(\frac{1}{3}{u^3} – u) + C = {3^3}(\frac{1}{3}{\sec ^3}\theta – \sec \theta ) + C = \\ \\ {3^3}\left[ {\frac{1}{3}\frac{{{{({x^2} + 9)}^{\frac{3}{2}}}}}{{{3^3}}} – \frac{{\sqrt {{x^2} + 9} }}{3}} \right] + C = \frac{1}{3}{({x^2} + 9)^{\frac{3}{2}}} – 9\sqrt {{x^2} + 9} + C \\ [/math]
سوال 3:
[math] \int {\frac{1}{{9{x^2} + 6x – 8}}} dx \\ \\9{x^2} + 6x – 8 = {(3x + 1)^2} – 9 \to u = 3x + 1,du = 3xdx \\ \\\int {\frac{1}{{9{x^2} + 6x – 8}}} dx = \int {\frac{{\frac{1}{3}du}}{{\sqrt {{u^2} – 9} }}} \\ u = 3\sec \theta \to du = 3\sec \theta \tan \thetad\theta ,\left[ {\sqrt {{u^2} – 9} = 3\tan \theta } \right] \\ \\ \int {\frac{{\frac{1}{3}du}}{{\sqrt {{u^2} – 9} }}} = \int {\frac{{\sec \theta \tan \theta d\theta }}{{3\tan \theta }}} = \frac{1}{3}\int {\sec \theta d\theta = } \frac{1}{3}\ln (\sec \theta + \tan \theta ) + {C_1} \\ \\ \\ = \frac{1}{3}\ln (u + \sqrt {{u^2} – 9} ) + C = \frac{1}{3}\ln (3x + 1 + \sqrt {9{x^2} + 6x – 8} ) + C \\[/math]
