## 問題1

## 問題1

1. 問題の内容

次の6つの不定積分を計算します。

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

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

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

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

(5) $\int \sin mx \cdot \cos nx dx$ (ただし、$m, n$ は自然数)

(6) $\int (e^x + e^{-x})^2 dx$

## 解き方の手順

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

$\sqrt{x} = t$ と置換すると、$x = t^2$, $dx = 2t dt$ より、

\begin{align*} \label{eq:1} \int \frac{(t+2)^3}{t} 2t dt &= 2 \int (t+2)^3 dt \\ &= 2 \int (t^3 + 6t^2 + 12t + 8) dt \\ &= 2 \left( \frac{1}{4} t^4 + 2t^3 + 6t^2 + 8t \right) + C \\ &= \frac{1}{2} x^2 + 4x\sqrt{x} + 12x + 16\sqrt{x} + C\end{align*}

(2) $\int \frac{x+\cos^2 x}{x \cos^2 x} dx = \int \left( \frac{1}{\cos^2 x} + \frac{1}{x} \right) dx = \int (\sec^2 x + \frac{1}{x}) dx = \tan x + \ln |x| + C$

(3) $\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \frac{1}{2} x + \frac{1}{4} \sin 2x + C$

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

$\int \sin^2 x dx = \int \frac{1-\cos 2x}{2} dx = \frac{1}{2} x - \frac{1}{4} \sin 2x + C$

$\int \csc^2 x dx = - \cot x + C$

よって、$\int (\sin^2 x + 2 + \csc^2 x) dx = \frac{1}{2} x - \frac{1}{4} \sin 2x + 2x - \cot x + C = \frac{5}{2} x - \frac{1}{4} \sin 2x - \cot x + C$

(5) $\int \sin mx \cos nx dx = \frac{1}{2} \int (\sin(m+n)x + \sin(m-n)x) dx$

$\frac{1}{2} \int (\sin(m+n)x + \sin(m-n)x) dx = \frac{1}{2} \left( - \frac{\cos(m+n)x}{m+n} - \frac{\cos(m-n)x}{m-n} \right) + C = - \frac{\cos(m+n)x}{2(m+n)} - \frac{\cos(m-n)x}{2(m-n)} + C$ （ただし、$m \neq n$）

もし$m = n$なら、

$\int \sin mx \cos mx dx = \frac{1}{2} \int \sin 2mx dx = - \frac{\cos 2mx}{4m} + C$

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

## 最終的な答え

(1) $\frac{1}{2} x^2 + 4x\sqrt{x} + 12x + 16\sqrt{x} + C$

(2) $\tan x + \ln |x| + C$

(3) $\frac{1}{2} x + \frac{1}{4} \sin 2x + C$

(4) $\frac{5}{2} x - \frac{1}{4} \sin 2x - \cot x + C$

(5) $- \frac{\cos(m+n)x}{2(m+n)} - \frac{\cos(m-n)x}{2(m-n)} + C$ ($m \neq n$のとき) または $- \frac{\cos 2mx}{4m} + C$ ($m=n$のとき)

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