【2023藤田医科大学・医学部】置換積分の計算

解答・解説

(１)　\(\displaystyle\int^{99}_{3}\sqrt{\sqrt{1+x}-1} dx\)

\(t=\sqrt{\sqrt{1+x}-1}\) とおくと

\((t^2+1)^2=1+x\) より

\(2(t^2+1)\cdot 2t\cdot dt=dx\)

よって，

\(\displaystyle\int^{99}_{3}\sqrt{\sqrt{1+x}-1} dx\)

\(=\displaystyle\int^{3}_{1}t\cdot 2(t^2+1)\cdot 2t dt\)

\(=\displaystyle\int^{3}_{1}4(t^4+t^2) dt\)

\(=4\Bigl[\displaystyle\frac{t^5}{5}+\displaystyle\frac{t^3}{3}\Bigr]^{3}_{1}\)

\(=\displaystyle\frac{3424}{15}\)

(２)　\(\displaystyle\int^{3}_{1}\sqrt{\displaystyle\frac{4}{x}-1} dx\)

\(x=4\cos^2 \theta\) \(\left(0≦\theta<\displaystyle\frac{\pi}{2}\right)\) とおくと

\(dx=-8\sin \theta\cos \theta dθ\) より

\(\displaystyle\int^{3}_{1}\sqrt{\displaystyle\frac{4}{x}-1} dx\)

\(=\displaystyle\int^{\frac{\pi}{6}}_{\frac{\pi}{3}}\sqrt{\displaystyle\frac{1}{\cos^2 \theta}-1}\cdot(-8)\sin \theta\cos \theta dθ\)

\(1+\tan^2 \theta=\displaystyle\frac{1}{\cos^2 \theta}\) より

(与式)\(=8\displaystyle\int^{\frac{\pi}{3}}_{\frac{\pi}{6}}\sqrt{\tan^2 \theta} \sin \theta\cos \theta dθ\)

\(\displaystyle\frac{\pi}{6}≦ \theta≦\displaystyle\frac{\pi}{3}\) のとき

\(\sqrt{\tan^2 \theta}=|\tan\theta|=\tan\theta=\displaystyle\frac{\sin \theta}{\cos \theta}\) より

(与式)\(=8\displaystyle\int^{\frac{\pi}{3}}_{\frac{\pi}{6}}\sin^2\theta dθ\)

半角の公式 \(\sin^2 \theta=\displaystyle\frac{1-\cos 2 \theta}{2}\) より

(与式)\(=4\displaystyle\int^{\frac{\pi}{3}}_{\frac{\pi}{6}}(1-\cos 2 \theta) dθ\)

\(=4\Bigl[ \theta-\displaystyle\frac{\sin 2 \theta}{2}\Bigr]^{\frac{\pi}{3}}_{\frac{\pi}{6}}\)

\(=\displaystyle\frac{2\pi}{3}\)