3.8 Answer

3.8

1.

(a) Let $t = \cos{x}$. Then $dt = -\sin{x}dx$．Note that

$$\begin{array}{llll} x&0 &\Rightarrow & \frac{\pi}{2}\\ \hline t&1 &\Rightarrow & 0 \end{array}$$

Then

$$\int_{0}^{\frac{\pi}{2}}{\cos^{4}{x}\sin{x}}\ dx = \int_{1}^{0}{t^{4}(-dt)}$$

$$= \int_{0}^{1}t^{4}\ dt = \frac{t^{5}}{5}\mid_{0}^{1} = \frac{1}{5}$$

(b) Let $t = \sqrt{1 + \cos{x}}$. Then $t^2 = 1 + \cos{x}$ and $2tdt = -\sin{x}dx$．Note that

$$\begin{array}{llll} x&0 &\Rightarrow & \frac{\pi}{2}\\ \hline t&\sqrt{2} &\Rightarrow & 1 \end{array}$$

Then

$$\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sqrt{1 + \cos{x}}}}\ dx = \int_{\sqrt{2}}^{1}{\frac{-2tdt}{t}}$$

$$= -2\int_{0}^{\frac{\pi}{2}}\ dt = -2t\mid_{\sqrt{2}}^{1} = -2(1 - \sqrt{2}) = 2(\sqrt{2} - 1)$$

(c)

$$\int_{0}^{\frac{\pi}{2}}{\sin^{4}{x}}\ dx = \frac{3!!}{4!!}\frac{\pi}{2} = \frac{3\pi}{16}$$

(d)

$$\int{f(x)g'(x)}\ dx = f(x)g(x) - \int{g(x)f'(x)}\ dx$$

Then

$$\begin{array}{ll} f(x) = x^2 & g'(x) = \cos{x}\\ f'(x) = 2x & g(x) = \sin{x} \end{array}$$

$$\int_{-1}^{1}{x^{2}\cos{x}}\ dx = x^{2}\sin{x}\mid_{-1}^{1} - \int{2x\sin{x}}\ dx$$

$$= \sin{1} - \sin{(-1)} - \int{2x\sin{x}}\ dx = 2\sin{1} - \int{2x\sin{x}}\ dx$$

Use the integration by parts again.

$$\begin{array}{ll} f(x) = 2x & g'(x) = \sin{x}\\ f'(x) = 2 & g(x) = -\cos{x} \end{array}$$

implies that

$$\int{2x\sin{x}}\ dx = -2x\cos{x}\mid_{-1}^{1} + 2\int{\cos{x}}\ dx$$

$$= -2\cos{1} - 2\cos{(-1)} + 2\sin{x}\mid_{-1}^{1}$$

$$= -2\cos{1} - 2\cos{1} + 2\sin{1} - 2\sin{(-1)} = -4\cos{1} + 4\sin{1}$$

Therefore，

$$\int_{-1}^{1}{x^{2}\cos{x}}\ dx = 2\sin{1} -[-4\cos{1} + 4\sin{1}]$$

$$= 4\cos{1} - 2\sin{1}$$

(e) For $n > 0$,

$$\int_{0}^{\pi}\cos{nx}\ dx = \frac{1}{n}\sin{nx}\mid_{0}^{\pi} = 0$$

2. $${I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx = \int_{0}^{\frac{\pi}{2}}\sin^{n-1}{x} \sin{x}dx = \frac{n-1}{n} I_{n-2}}$$

$${J_{n} = \int_{0}^{\frac{\pi}{2}}\cos^{n}{x}dx = \int_{0}^{\frac{\pi}{2}}\cos^{n-1}{x} \cos{x}dx = \frac{n-1}{n} J_{n-2}}$$