高専の数学解説

第4章 4.17

(1)

1−x2x2\frac{\sqrt{1-x^2}}{x^2}

∫1−x2x2 dx\displaystyle\int \frac{\sqrt{1-x^2}}{x^2}\,dx x=sin⁡tx=\sin t とおくと dx=cos⁡t dtdx=\cos t\,dt これから

=∫cos⁡tsin⁡2t⋅cos⁡t dt=∫cos⁡2tsin⁡2t dt=∫1−sin⁡2tsin⁡2t dt=∫1sin⁡2t−1 dt=−1tan⁡t−t+C=\int \frac{\cos t}{\sin^2 t}\cdot\cos t\,dt=\int \frac{\cos^2 t}{\sin^2 t}\,dt=\int \frac{1-\sin^2 t}{\sin^2 t}\,dt=\int \frac{1}{\sin^2 t}-1\,dt=-\frac{1}{\tan t}-t+C 元に戻して

=−1tan⁡(sin⁡−1x)−sin⁡−1x+C=−1x1−x2−sin⁡−1x+C=−1−x2x−sin⁡−1x+C=-\frac{1}{\tan(\sin^{-1} x)}-\sin^{-1}x+C=-\cfrac{1}{\dfrac{x}{\sqrt{1-x^2}}}-\sin^{-1}x+C=-\frac{\sqrt{1-x^2}}{x}-\sin^{-1}x+C （C は積分定数）

(2)

1x−2x+2x−3\frac{1}{x-2}\sqrt{\frac{x+2}{x-3}}

∫1x−2x+2x−3 dx\displaystyle\int \frac{1}{x-2}\sqrt{\frac{x+2}{x-3}}\,dx x+2x−3=t\sqrt{\dfrac{x+2}{x-3}}=t とおくと x=3t2+2t2−1x=\dfrac{3t^2+2}{t^2-1}

dx=6t(t2−1)−2t(3t2+2)(t2−1)2 dt=−10t(t2−1)2 dtdx=\frac{6t(t^2-1)-2t(3t^2+2)}{(t^2-1)^2}\,dt=\frac{-10t}{(t^2-1)^2}\,dt これから

=∫t3t2+2t2−1−2⋅−10t(t2−1)2 dt=∫t(t2−1)3t2+2−2t2+2⋅−10t(t2−1)2 dt=∫−10t2(t2+4)(t2−1) dt=\int \frac{t}{\dfrac{3t^2+2}{t^2-1}-2}\cdot\frac{-10t}{(t^2-1)^2}\,dt=\int \frac{t(t^2-1)}{3t^2+2-2t^2+2}\cdot\frac{-10t}{(t^2-1)^2}\,dt=\int \frac{-10t^2}{(t^2+4)(t^2-1)}\,dt

=∫−8t2+4+−2t2−1 dt=−4tan⁡−1t2−2log⁡∣t−1t+1∣+C=\int \frac{-8}{t^2+4}+\frac{-2}{t^2-1}\,dt=-4\tan^{-1}\frac{t}{2}-2\log\left|\frac{t-1}{t+1}\right|+C 元に戻して

=−4tan⁡−112x+2x−3−2log⁡∣x+2x−3−1x+2x−3+1∣+C=-4\tan^{-1}\frac{1}{2}\sqrt{\frac{x+2}{x-3}}-2\log\left|\dfrac{\sqrt{\dfrac{x+2}{x-3}}-1}{\sqrt{\dfrac{x+2}{x-3}}+1}\right|+C

−4tan⁡−112x+2x−3−2log⁡∣x+2−x−3∣+C-4\tan^{-1}\frac{1}{2}\sqrt{\frac{x+2}{x-3}}-2\log\left|\sqrt{x+2}-\sqrt{x-3}\right|+C （C は積分定数）

解説: r31bn1z