問題は以下の４つのパートに分かれています。 (1) 三角関数の値を求める問題 (2) 対数の計算問題 (3) 関数の微分問題 (4) 不定積分/定積分の計算問題

はい、承知いたしました。問題用紙に記載された問題を解きます。

1. 問題の内容

問題は以下の４つのパートに分かれています。

(1) 三角関数の値を求める問題

(2) 対数の計算問題

(3) 関数の微分問題

(4) 不定積分/定積分の計算問題

2. 解き方の手順

**

1. 三角関数の値を求める問題**

(1) $\sin(-\frac{5\pi}{6})$

$\sin(-\frac{5\pi}{6}) = -\sin(\frac{5\pi}{6}) = -\sin(\pi - \frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}$

(2) $\cos(\frac{3\pi}{4})$

$\cos(\frac{3\pi}{4}) = \cos(\pi - \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$

(3) $\tan(\frac{7\pi}{6})$

$\tan(\frac{7\pi}{6}) = \tan(\pi + \frac{\pi}{6}) = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

**

2. 対数の計算問題**

(1) $\log_{2\sqrt{2}} 27 + 3\log_2 \frac{4}{3}$

$\log_{2\sqrt{2}} 27 = \log_{2^{3/2}} 3^3 = \frac{3}{3/2}\log_2 3 = 2\log_2 3$

$3\log_2 \frac{4}{3} = 3(\log_2 4 - \log_2 3) = 3(2-\log_2 3) = 6 - 3\log_2 3$

したがって、

$\log_{2\sqrt{2}} 27 + 3\log_2 \frac{4}{3} = 2\log_2 3 + 6 - 3\log_2 3 = 6 - \log_2 3$

(2) $2\log_{10} \frac{\sqrt{3}}{10} - \log_{10} 30$

$2\log_{10} \frac{\sqrt{3}}{10} = 2(\log_{10} \sqrt{3} - \log_{10} 10) = 2(\frac{1}{2}\log_{10} 3 - 1) = \log_{10} 3 - 2$

$\log_{10} 30 = \log_{10} (3\times 10) = \log_{10} 3 + \log_{10} 10 = \log_{10} 3 + 1$

したがって、

$2\log_{10} \frac{\sqrt{3}}{10} - \log_{10} 30 = \log_{10} 3 - 2 - (\log_{10} 3 + 1) = -3$

**

3. 関数の微分問題**

(1) $y=x^4$

$y' = 4x^3$

(2) $y = 3\sqrt{x}$

$y' = 3\cdot \frac{1}{2\sqrt{x}} = \frac{3}{2\sqrt{x}}$

(3) $y = \cos x$

$y' = -\sin x$

(4) $y = x^3\sin x$

$y' = 3x^2\sin x + x^3\cos x$

(5) $y = \tan x + e^x$

$y' = \frac{1}{\cos^2 x} + e^x = \sec^2 x + e^x$

(6) $y = \cos x^2$

$y' = -\sin x^2 \cdot 2x = -2x\sin x^2$

(7) $y = (x^2+1)^{-1/3}$

$y' = -\frac{1}{3}(x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}$

(8) $y = x^x$

$\log y = x \log x$

$\frac{y'}{y} = \log x + x \cdot \frac{1}{x} = \log x + 1$

$y' = y(\log x + 1) = x^x(\log x + 1)$

(9) $y = xe^{2x+3}$

$y' = e^{2x+3} + x e^{2x+3} \cdot 2 = e^{2x+3} + 2xe^{2x+3} = (1+2x)e^{2x+3}$

**

4. 不定積分/定積分の計算問題**

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

(2) $\int \cos x dx = \sin x + C$

(3) $\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C = (x-1)e^x + C$

(4) $\int 2\sqrt{x} dx = 2 \int x^{1/2} dx = 2 \cdot \frac{2}{3}x^{3/2} + C = \frac{4}{3}x^{3/2} + C$

(5) $\int \frac{1}{x\log x} dx = \int \frac{1}{\log x} \cdot \frac{1}{x} dx = \int \frac{1}{\log x} d(\log x) = \log |\log x| + C$

(6) $\int \sqrt{2x-3} dx = \int (2x-3)^{1/2} dx = \frac{1}{2} \int (2x-3)^{1/2} d(2x-3) = \frac{1}{2} \cdot \frac{2}{3}(2x-3)^{3/2} + C = \frac{1}{3}(2x-3)^{3/2} + C$

(7) $\int_{-1}^2 4x dx = [2x^2]_{-1}^2 = 2(2^2) - 2(-1)^2 = 8 - 2 = 6$

(8) $\int_0^{\pi} \sin x dx = [-\cos x]_0^{\pi} = -\cos \pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2$

(9) $\int_1^e \frac{(\log x)^2}{x} dx = \int_1^e (\log x)^2 d(\log x) = [\frac{1}{3}(\log x)^3]_1^e = \frac{1}{3}(\log e)^3 - \frac{1}{3}(\log 1)^3 = \frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 = \frac{1}{3}$

3. 最終的な答え

1. (1) $-\frac{1}{2}$ (2) $-\frac{\sqrt{2}}{2}$ (3) $\frac{\sqrt{3}}{3}$

2. (1) $6-\log_2 3$ (2) $-3$

3. (1) $4x^3$ (2) $\frac{3}{2\sqrt{x}}$ (3) $-\sin x$ (4) $3x^2\sin x + x^3\cos x$ (5) $\sec^2 x + e^x$ (6) $-2x\sin x^2$ (7) $-\frac{2x}{3(x^2+1)^{4/3}}$ (8) $x^x(\log x + 1)$ (9) $(1+2x)e^{2x+3}$

4. (1) $\frac{1}{3}x^3 + x + C$ (2) $\sin x + C$ (3) $(x-1)e^x + C$ (4) $\frac{4}{3}x^{3/2} + C$ (5) $\log |\log x| + C$ (6) $\frac{1}{3}(2x-3)^{3/2} + C$ (7) $6$ (8) $2$ (9) $\frac{1}{3}$