5つの積分問題を解きます。 (1) $\int (4-5 \tan x) \cos x dx$ (2) $\int \frac{dx}{\sqrt[3]{(1-3x)^2}}$ (3) $\int x^3 \sqrt{4-x^2} dx$ (4) $\int e^{-x} \sin(x+\frac{\pi}{4}) dx$ (5) $\int (3x-4)^2 dx$

1. 問題の内容

5つの積分問題を解きます。

(1) $\int (4-5 \tan x) \cos x dx$

(2) $\int \frac{dx}{\sqrt[3]{(1-3x)^2}}$

(3) $\int x^3 \sqrt{4-x^2} dx$

(4) $\int e^{-x} \sin(x+\frac{\pi}{4}) dx$

(5) $\int (3x-4)^2 dx$

2. 解き方の手順

(1) $\int (4-5 \tan x) \cos x dx$

$=\int (4\cos x - 5 \frac{\sin x}{\cos x} \cos x) dx$

$=\int (4\cos x - 5 \sin x) dx$

$=4\int \cos x dx - 5\int \sin x dx$

$=4\sin x - 5(-\cos x) + C$

$=4\sin x + 5\cos x + C$

(2) $\int \frac{dx}{\sqrt[3]{(1-3x)^2}}$

$=\int (1-3x)^{-2/3} dx$

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

$\int (1-3x)^{-2/3} dx = \int u^{-2/3} (-\frac{1}{3}) du = -\frac{1}{3} \int u^{-2/3} du$

$=-\frac{1}{3} \frac{u^{1/3}}{1/3} + C = -u^{1/3} + C = -(1-3x)^{1/3} + C$

$=-\sqrt[3]{1-3x} + C$

(3) $\int x^3 \sqrt{4-x^2} dx$

$u = 4-x^2$ と置換すると、 $du = -2x dx$, $x^2 = 4-u$, $x dx = -\frac{1}{2} du$

$\int x^3 \sqrt{4-x^2} dx = \int x^2 \sqrt{4-x^2} x dx = \int (4-u) \sqrt{u} (-\frac{1}{2}) du = -\frac{1}{2} \int (4-u) u^{1/2} du$

$=-\frac{1}{2} \int (4u^{1/2} - u^{3/2}) du = -\frac{1}{2} (4 \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2}) + C = -\frac{1}{2} (\frac{8}{3} u^{3/2} - \frac{2}{5} u^{5/2}) + C$

$=-\frac{4}{3} u^{3/2} + \frac{1}{5} u^{5/2} + C = -\frac{4}{3} (4-x^2)^{3/2} + \frac{1}{5} (4-x^2)^{5/2} + C$

$= (4-x^2)^{3/2} (-\frac{4}{3} + \frac{1}{5} (4-x^2)) + C = (4-x^2)^{3/2} (-\frac{20}{15} + \frac{12}{15} - \frac{x^2}{5}) + C$

$= (4-x^2)^{3/2} (-\frac{8}{15} - \frac{x^2}{5}) + C = -\frac{1}{15} (4-x^2)^{3/2} (8 + 3x^2) + C$

(4) $\int e^{-x} \sin(x+\frac{\pi}{4}) dx$

$\sin(x+\frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} (\sin x + \cos x)$

$\int e^{-x} \sin(x+\frac{\pi}{4}) dx = \frac{\sqrt{2}}{2} \int e^{-x} (\sin x + \cos x) dx = \frac{\sqrt{2}}{2} (\int e^{-x} \sin x dx + \int e^{-x} \cos x dx)$

$I = \int e^{-x} \sin x dx$ , $J = \int e^{-x} \cos x dx$

$I = -e^{-x} \sin x + \int e^{-x} \cos x dx = -e^{-x} \sin x + J$

$J = -e^{-x} \cos x - \int e^{-x} \sin x dx = -e^{-x} \cos x - I$

$I = -e^{-x} \sin x - e^{-x} \cos x - I$

$2I = -e^{-x} (\sin x + \cos x)$

$I = -\frac{1}{2} e^{-x} (\sin x + \cos x)$

$J = -e^{-x} \cos x - (-\frac{1}{2} e^{-x} (\sin x + \cos x)) = -e^{-x} \cos x + \frac{1}{2} e^{-x} (\sin x + \cos x) = \frac{1}{2} e^{-x} (\sin x - \cos x)$

$\int e^{-x} \sin(x+\frac{\pi}{4}) dx = \frac{\sqrt{2}}{2} (-\frac{1}{2} e^{-x} (\sin x + \cos x) + \frac{1}{2} e^{-x} (\sin x - \cos x)) + C$

$= \frac{\sqrt{2}}{2} \frac{1}{2} e^{-x} (-2 \cos x) + C = -\frac{\sqrt{2}}{2} e^{-x} \cos x + C$

(5) $\int (3x-4)^2 dx$

$=\int (9x^2 - 24x + 16) dx = 9\int x^2 dx - 24 \int x dx + 16 \int dx$

$=9 \frac{x^3}{3} - 24 \frac{x^2}{2} + 16x + C = 3x^3 - 12x^2 + 16x + C$

3. 最終的な答え

(1) $4\sin x + 5\cos x + C$

(2) $-\sqrt[3]{1-3x} + C$

(3) $-\frac{1}{15} (4-x^2)^{3/2} (8 + 3x^2) + C$

(4) $-\frac{\sqrt{2}}{2} e^{-x} \cos x + C$

(5) $3x^3 - 12x^2 + 16x + C$