与えられた9つの不定積分を計算する問題です。 (1) $\int \frac{dx}{\cos x}$ ($\sin x = t$ と置換) (2) $\int \sin^4 x \cos^3 x dx$ (3) $\int \frac{dx}{1 + \sin x}$ (4) $\int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$ ($ab \neq 0$) (5) $\int \tan^4 x dx$ (6) $\int \frac{dx}{\sin^4 x}$ (7) $\int \frac{dx}{5 + 3 \sin x + 4 \cos x}$ (8) $\int \frac{3}{\tan x (2 + \cos^2 x)} dx$ (9) $\int \frac{dx}{5 + 3 \sin x}$

1. 問題の内容

与えられた9つの不定積分を計算する問題です。

(1) $\int \frac{dx}{\cos x}$ ($\sin x = t$ と置換)

(2) $\int \sin^4 x \cos^3 x dx$

(3) $\int \frac{dx}{1 + \sin x}$

(4) $\int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$ ($ab \neq 0$)

(5) $\int \tan^4 x dx$

(6) $\int \frac{dx}{\sin^4 x}$

(7) $\int \frac{dx}{5 + 3 \sin x + 4 \cos x}$

(8) $\int \frac{3}{\tan x (2 + \cos^2 x)} dx$

(9) $\int \frac{dx}{5 + 3 \sin x}$

2. 解き方の手順

(1) $\int \frac{dx}{\cos x}$ ($\sin x = t$ と置換)

$\sin x = t$ と置換すると、$\cos x dx = dt$ より $dx = \frac{dt}{\cos x}$。

$\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - t^2}$ なので、

$\int \frac{dx}{\cos x} = \int \frac{1}{\cos x} \frac{dt}{\cos x} = \int \frac{dt}{\cos^2 x} = \int \frac{dt}{1 - t^2} = \int \frac{dt}{(1 - t)(1 + t)}$.

部分分数分解すると $\frac{1}{1 - t^2} = \frac{1}{2} \left( \frac{1}{1 - t} + \frac{1}{1 + t} \right)$.

したがって、$\int \frac{dt}{1 - t^2} = \frac{1}{2} \int \left( \frac{1}{1 - t} + \frac{1}{1 + t} \right) dt = \frac{1}{2} (-\ln|1 - t| + \ln|1 + t|) + C = \frac{1}{2} \ln \left| \frac{1 + t}{1 - t} \right| + C = \frac{1}{2} \ln \left| \frac{1 + \sin x}{1 - \sin x} \right| + C = \ln \left| \frac{1 + \sin x}{\cos x} \right| + C = \ln |\sec x + \tan x| + C$.

(2) $\int \sin^4 x \cos^3 x dx$

$\cos^3 x = \cos^2 x \cos x = (1 - \sin^2 x) \cos x$ なので、

$\int \sin^4 x \cos^3 x dx = \int \sin^4 x (1 - \sin^2 x) \cos x dx$.

$t = \sin x$ とおくと $dt = \cos x dx$。

$\int t^4 (1 - t^2) dt = \int (t^4 - t^6) dt = \frac{t^5}{5} - \frac{t^7}{7} + C = \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C$.

(3) $\int \frac{dx}{1 + \sin x}$

$\int \frac{dx}{1 + \sin x} = \int \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx = \int \frac{1 - \sin x}{1 - \sin^2 x} dx = \int \frac{1 - \sin x}{\cos^2 x} dx = \int (\sec^2 x - \sec x \tan x) dx = \tan x - \sec x + C$.

(4) $\int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$

$\int \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \int \frac{1}{\cos^2 x} \frac{dx}{a^2 + b^2 \tan^2 x} = \int \frac{\sec^2 x}{a^2 + b^2 \tan^2 x} dx$.

$t = \tan x$ とおくと $dt = \sec^2 x dx$。

$\int \frac{dt}{a^2 + b^2 t^2} = \frac{1}{b^2} \int \frac{dt}{(a/b)^2 + t^2} = \frac{1}{b^2} \cdot \frac{1}{a/b} \arctan \frac{t}{a/b} + C = \frac{1}{ab} \arctan \frac{b \tan x}{a} + C$.

(5) $\int \tan^4 x dx$

$\int \tan^4 x dx = \int \tan^2 x \tan^2 x dx = \int \tan^2 x (\sec^2 x - 1) dx = \int \tan^2 x \sec^2 x dx - \int \tan^2 x dx$.

$\int \tan^2 x \sec^2 x dx = \int t^2 dt = \frac{t^3}{3} = \frac{\tan^3 x}{3}$ ($t = \tan x$ と置換)。

$\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x$.

$\int \tan^4 x dx = \frac{\tan^3 x}{3} - (\tan x - x) + C = \frac{\tan^3 x}{3} - \tan x + x + C$.

(6) $\int \frac{dx}{\sin^4 x}$

$\int \frac{dx}{\sin^4 x} = \int \csc^4 x dx = \int \csc^2 x (1 + \cot^2 x) dx = \int \csc^2 x dx + \int \csc^2 x \cot^2 x dx$.

$\int \csc^2 x dx = - \cot x$.

$\int \csc^2 x \cot^2 x dx = - \frac{\cot^3 x}{3}$.

$\int \frac{dx}{\sin^4 x} = - \cot x - \frac{\cot^3 x}{3} + C$.

(7) $\int \frac{dx}{5 + 3 \sin x + 4 \cos x}$

$t = \tan(x/2)$ とおくと、$\sin x = \frac{2t}{1 + t^2}$, $\cos x = \frac{1 - t^2}{1 + t^2}$, $dx = \frac{2}{1 + t^2} dt$.

$\int \frac{dx}{5 + 3 \sin x + 4 \cos x} = \int \frac{1}{5 + 3 \frac{2t}{1 + t^2} + 4 \frac{1 - t^2}{1 + t^2}} \frac{2}{1 + t^2} dt = \int \frac{2}{5(1 + t^2) + 6t + 4(1 - t^2)} dt = \int \frac{2}{t^2 + 6t + 9} dt = \int \frac{2}{(t + 3)^2} dt = - \frac{2}{t + 3} + C = - \frac{2}{\tan(x/2) + 3} + C$.

(8) $\int \frac{3}{\tan x (2 + \cos^2 x)} dx$

$\int \frac{3}{\tan x (2 + \cos^2 x)} dx = \int \frac{3 \cos x}{\sin x (2 + \cos^2 x)} dx = \int \frac{3 \cos x}{\sin x (2 + 1 - \sin^2 x)} dx = \int \frac{3 \cos x}{\sin x (3 - \sin^2 x)} dx$.

$t = \sin x$ とおくと $dt = \cos x dx$.

$\int \frac{3}{t (3 - t^2)} dt = \int \left( \frac{1}{t} + \frac{t}{3 - t^2} \right) dt = \ln|t| - \frac{1}{2} \ln|3 - t^2| + C = \ln|\sin x| - \frac{1}{2} \ln|3 - \sin^2 x| + C$.

(9) $\int \frac{dx}{5 + 3 \sin x}$

$t = \tan(x/2)$ とおくと、$\sin x = \frac{2t}{1 + t^2}$, $dx = \frac{2}{1 + t^2} dt$.

$\int \frac{dx}{5 + 3 \sin x} = \int \frac{1}{5 + 3 \frac{2t}{1 + t^2}} \frac{2}{1 + t^2} dt = \int \frac{2}{5(1 + t^2) + 6t} dt = \int \frac{2}{5t^2 + 6t + 5} dt = \frac{2}{5} \int \frac{dt}{t^2 + \frac{6}{5} t + 1} = \frac{2}{5} \int \frac{dt}{(t + \frac{3}{5})^2 + 1 - \frac{9}{25}} = \frac{2}{5} \int \frac{dt}{(t + \frac{3}{5})^2 + \frac{16}{25}} = \frac{2}{5} \cdot \frac{5}{4} \arctan \left( \frac{t + \frac{3}{5}}{\frac{4}{5}} \right) + C = \frac{1}{2} \arctan \left( \frac{5t + 3}{4} \right) + C = \frac{1}{2} \arctan \left( \frac{5 \tan(x/2) + 3}{4} \right) + C$.

3. 最終的な答え

(1) $\ln |\sec x + \tan x| + C$

(2) $\frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C$

(3) $\tan x - \sec x + C$

(4) $\frac{1}{ab} \arctan \frac{b \tan x}{a} + C$

(5) $\frac{\tan^3 x}{3} - \tan x + x + C$

(6) $- \cot x - \frac{\cot^3 x}{3} + C$

(7) $- \frac{2}{\tan(x/2) + 3} + C$

(8) $\ln|\sin x| - \frac{1}{2} \ln|3 - \sin^2 x| + C$

(9) $\frac{1}{2} \arctan \left( \frac{5 \tan(x/2) + 3}{4} \right) + C$