第8章 8.6

(1)

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

(2)

$\int (4x+5)^5 dx = \frac{1}{4}\cdot\frac{1}{6}(4x+5)^6 = \frac{1}{24}(4x+5)^6$

(3)

$\int \frac{2x+4}{x^2-1}dx = \int \left(\frac{3}{x-1}-\frac{1}{x+1}\right)dx = \log|x-1|^3 - \log|x+1| = \log\frac{|x-1|^3}{|x+1|}$

(4)

$\displaystyle\int \sin^2 x \cdot \cos x\,dx$ $\sin x = t$ とおくと $\cos x\,dx = dt$ これより

$\displaystyle\int t^2 dt = \frac{1}{3}t^3$ 元に戻して $\displaystyle\int \sin^2 x \cdot \cos x\,dx = \frac{1}{3}\sin^3 x$

(5)

$\displaystyle\int \frac{\cos x}{2+\sin x}dx$ $\sin x = t$ とおくと $\cos x\,dx = dt$ これより

$\displaystyle\int \frac{1}{2+t}dt = \log|2+t|$ 元に戻して $\displaystyle\int \frac{\cos x}{2+\sin x}dx = \log(2+\sin x)$

(6)

$\displaystyle\int \frac{e^x}{e^{2x}-4}dx$ $e^x = t$ とおくと $e^x dx = dt$ これより

$\displaystyle\int \frac{1}{t^2-4}dt = \int \frac{1}{4}\left(\frac{1}{t-2}-\frac{1}{t+2}\right)dt = \frac{1}{4}\log\frac{|t-2|}{|t+2|}$ 元に戻して $\displaystyle\int \frac{e^x}{e^{2x}-4}dx = \frac{1}{4}\log\frac{|e^x-2|}{|e^x+2|}$

(7)

$\int \sin 2x \cdot \cos 3x\, dx = \int \frac{1}{2}\{\sin 5x - \sin x\}dx = \frac{1}{2}\left(-\frac{1}{5}\cos 5x + \cos x\right)$

(8)

$\int \sin^2 x \cos^2 x\, dx = \int (1-\cos^2 x)\cos^2 x\, dx = \int \cos^2 x - \cos^4 x\, dx = \int \frac{1+\cos 2x}{2} - \left(\frac{1+\cos 2x}{2}\right)^2 dx$

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

$= \int \left(\frac{1}{8}-\frac{1}{8}\cos 4x\right)dx = \frac{1}{8}x - \frac{1}{32}\sin 4x = \frac{1}{8}\left(x-\frac{1}{4}\sin 4x\right)$