与えられた関数の不定積分を計算する問題です。具体的には、以下の7つの不定積分を求める必要があります。 (1) $\int \frac{1}{\sqrt{1-9x^2}} dx$ (2) $\int \frac{1}{\sqrt{9-x^2}} dx$ (3) $\int \frac{1}{1+9x^2} dx$ (4) $\int \frac{1}{9+x^2} dx$ (5) $\int \frac{1}{x^2-2x+5} dx$ (6) $\int \frac{1}{\sqrt{-x^2-6x-8}} dx$ (4) $\int \frac{x}{1+x^2} dx$ (5) $\int \frac{x-1}{x^2+2x+6} dx$ (6) $\int \frac{\sin x \cos x}{\cos^2 x - 2} dx$ (7) $\int \tan x dx$

1. 問題の内容

与えられた関数の不定積分を計算する問題です。具体的には、以下の7つの不定積分を求める必要があります。

(1) $\int \frac{1}{\sqrt{1-9x^2}} dx$

(2) $\int \frac{1}{\sqrt{9-x^2}} dx$

(3) $\int \frac{1}{1+9x^2} dx$

(4) $\int \frac{1}{9+x^2} dx$

(5) $\int \frac{1}{x^2-2x+5} dx$

(6) $\int \frac{1}{\sqrt{-x^2-6x-8}} dx$

(4) $\int \frac{x}{1+x^2} dx$

(5) $\int \frac{x-1}{x^2+2x+6} dx$

(6) $\int \frac{\sin x \cos x}{\cos^2 x - 2} dx$

(7) $\int \tan x dx$

2. 解き方の手順

(1) $\int \frac{1}{\sqrt{1-9x^2}} dx$

$x = \frac{1}{3}\sin{\theta}$ と置換します。すると、$dx = \frac{1}{3}\cos{\theta} d\theta$ となり、

$\int \frac{1}{\sqrt{1-9x^2}} dx = \int \frac{1}{\sqrt{1-\sin^2{\theta}}} \cdot \frac{1}{3}\cos{\theta} d\theta = \int \frac{1}{\sqrt{\cos^2{\theta}}} \cdot \frac{1}{3}\cos{\theta} d\theta = \int \frac{1}{\cos{\theta}} \cdot \frac{1}{3}\cos{\theta} d\theta = \int \frac{1}{3} d\theta = \frac{1}{3}\theta + C = \frac{1}{3}\arcsin(3x) + C$

(2) $\int \frac{1}{\sqrt{9-x^2}} dx$

$x = 3\sin{\theta}$ と置換します。すると、$dx = 3\cos{\theta} d\theta$ となり、

$\int \frac{1}{\sqrt{9-x^2}} dx = \int \frac{1}{\sqrt{9-9\sin^2{\theta}}} \cdot 3\cos{\theta} d\theta = \int \frac{1}{3\sqrt{1-\sin^2{\theta}}} \cdot 3\cos{\theta} d\theta = \int \frac{1}{3\cos{\theta}} \cdot 3\cos{\theta} d\theta = \int 1 d\theta = \theta + C = \arcsin(\frac{x}{3}) + C$

(3) $\int \frac{1}{1+9x^2} dx$

$x = \frac{1}{3}\tan{\theta}$ と置換します。すると、$dx = \frac{1}{3}\sec^2{\theta} d\theta$ となり、

$\int \frac{1}{1+9x^2} dx = \int \frac{1}{1+\tan^2{\theta}} \cdot \frac{1}{3}\sec^2{\theta} d\theta = \int \frac{1}{\sec^2{\theta}} \cdot \frac{1}{3}\sec^2{\theta} d\theta = \int \frac{1}{3} d\theta = \frac{1}{3}\theta + C = \frac{1}{3}\arctan(3x) + C$

(4) $\int \frac{1}{9+x^2} dx$

$\int \frac{1}{9+x^2} dx = \frac{1}{9} \int \frac{1}{1+(\frac{x}{3})^2} dx$

$u = \frac{x}{3}$ と置換すると、$du = \frac{1}{3} dx$ なので、$dx = 3du$

$\frac{1}{9} \int \frac{1}{1+u^2} \cdot 3 du = \frac{1}{3} \int \frac{1}{1+u^2} du = \frac{1}{3} \arctan(u) + C = \frac{1}{3}\arctan(\frac{x}{3}) + C$

(5) $\int \frac{1}{x^2-2x+5} dx$

$x^2-2x+5 = (x-1)^2 + 4$ なので、

$\int \frac{1}{x^2-2x+5} dx = \int \frac{1}{(x-1)^2+4} dx$

$u = x-1$ と置換すると、$du = dx$

$\int \frac{1}{u^2+4} du = \frac{1}{4} \int \frac{1}{1+(\frac{u}{2})^2} du$

$v = \frac{u}{2}$ と置換すると、$dv = \frac{1}{2} du$ なので、$du = 2dv$

$\frac{1}{4} \int \frac{1}{1+v^2} \cdot 2dv = \frac{1}{2} \int \frac{1}{1+v^2} dv = \frac{1}{2}\arctan(v) + C = \frac{1}{2}\arctan(\frac{u}{2}) + C = \frac{1}{2}\arctan(\frac{x-1}{2}) + C$

(6) $\int \frac{1}{\sqrt{-x^2-6x-8}} dx$

$-x^2-6x-8 = -(x^2+6x+8) = -(x^2+6x+9-1) = -(x+3)^2+1 = 1-(x+3)^2$

$\int \frac{1}{\sqrt{-x^2-6x-8}} dx = \int \frac{1}{\sqrt{1-(x+3)^2}} dx$

$u = x+3$ と置換すると、$du = dx$

$\int \frac{1}{\sqrt{1-u^2}} du = \arcsin(u) + C = \arcsin(x+3) + C$

(4) $\int \frac{x}{1+x^2} dx$

$u = 1+x^2$ と置換すると、$du = 2x dx$ なので、$x dx = \frac{1}{2} du$

$\int \frac{x}{1+x^2} dx = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(1+x^2) + C$ (∵ $1+x^2 > 0$)

(5) $\int \frac{x-1}{x^2+2x+6} dx$

$\int \frac{x-1}{x^2+2x+6} dx = \int \frac{x-1}{(x+1)^2+5} dx$

$u = x+1$ と置換すると、$x = u-1$ なので、$x-1 = u-2$、$du = dx$

$\int \frac{u-2}{u^2+5} du = \int \frac{u}{u^2+5} du - \int \frac{2}{u^2+5} du$

$\int \frac{u}{u^2+5} du = \frac{1}{2} \ln(u^2+5) + C_1$

$\int \frac{2}{u^2+5} du = \frac{2}{5} \int \frac{1}{(\frac{u}{\sqrt{5}})^2+1} du$

$v = \frac{u}{\sqrt{5}}$ と置換すると、$dv = \frac{1}{\sqrt{5}} du$ なので、$du = \sqrt{5} dv$

$\frac{2}{5} \int \frac{1}{v^2+1} \sqrt{5} dv = \frac{2}{\sqrt{5}} \arctan(v) + C_2 = \frac{2}{\sqrt{5}} \arctan(\frac{u}{\sqrt{5}}) + C_2$

$\int \frac{x-1}{x^2+2x+6} dx = \frac{1}{2} \ln(u^2+5) - \frac{2}{\sqrt{5}} \arctan(\frac{u}{\sqrt{5}}) + C = \frac{1}{2} \ln(x^2+2x+6) - \frac{2}{\sqrt{5}} \arctan(\frac{x+1}{\sqrt{5}}) + C$

(6) $\int \frac{\sin x \cos x}{\cos^2 x - 2} dx$

$u = \cos^2 x - 2$ と置換すると、$du = -2\cos x \sin x dx$ なので、$\sin x \cos x dx = -\frac{1}{2} du$

$\int \frac{\sin x \cos x}{\cos^2 x - 2} dx = \int \frac{1}{u} (-\frac{1}{2} du) = -\frac{1}{2} \int \frac{1}{u} du = -\frac{1}{2} \ln|u| + C = -\frac{1}{2} \ln|\cos^2 x - 2| + C = -\frac{1}{2} \ln(2 - \cos^2 x) + C$ (∵ $2 - \cos^2 x > 0$)

(7) $\int \tan x dx$

$\int \tan x dx = \int \frac{\sin x}{\cos x} dx$

$u = \cos x$ と置換すると、$du = -\sin x dx$ なので、$\sin x dx = -du$

$\int \frac{\sin x}{\cos x} dx = \int \frac{1}{u} (-du) = -\int \frac{1}{u} du = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C$

3. 最終的な答え

(1) $\frac{1}{3}\arcsin(3x) + C$

(2) $\arcsin(\frac{x}{3}) + C$

(3) $\frac{1}{3}\arctan(3x) + C$

(4) $\frac{1}{3}\arctan(\frac{x}{3}) + C$

(5) $\frac{1}{2}\arctan(\frac{x-1}{2}) + C$

(6) $\arcsin(x+3) + C$

(4) $\frac{1}{2} \ln(1+x^2) + C$

(5) $\frac{1}{2} \ln(x^2+2x+6) - \frac{2}{\sqrt{5}} \arctan(\frac{x+1}{\sqrt{5}}) + C$

(6) $-\frac{1}{2} \ln(2 - \cos^2 x) + C$

(7) $\ln|\sec x| + C$