3.3 Answer

3.3

1.

(a) Set $u = x$ will not change anything．So, we let $u = \log{x}$. Then

$$\begin{array}{ll} u = \log{x}, & dv = x dx\begin{rawhtml} とおく... ...}{x}dx, & v = \int dv = \int x dx = \frac{x^2}{2} \end{array}$$

Thus,

$$\int \underbrace{\log{x}}_{u} \underbrace{x dx}_{dv} = \underbrace{\frac{x^2}{2}}_{v} \underbrace{\log{x}}_{u} - \int \underbrace{\frac{x^2}{2}}_{v} \underbrace{\frac{1}{x} dx}_{du}$$

$$= \frac{x^2 \log{x}}{2} - \int \frac{x}{2}dx = \frac{x^2 \log{x}}{2} - \frac{x^2}{4} + c$$

(b) Set

$$\begin{array}{ll} u = x^2, & dv = e^{-x} dx. \\ du = 2x dx, & v = \int dv = \int e^{-x} dx = - e^{-x} \end{array}$$

Then

$$\int x^2 e^{-x} dx = -x^2 e^{-x} + 2 \int x e^{-x} dx$$

We use the integration by parts one more time. Set

$$\begin{array}{ll} u = x, & dv = e^{-x} dx\\ du = dx, & v = \int dv = \int e^{-x} dx = - e^{-x} \end{array}$$

Then

$$\int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx = -x e^{-x} - e^{-x} + c$$

Therefore,

$$\int x^2 e^{-x} dx = \frac{-2 - 2\,x - {x^2}}{{e^x}}$$

(c)

Set

$$\begin{array}{ll} u = (\log{x})^2, & dv = dx\\ du = 2\log{x}\frac{1}{x} dx, & v = \int dv = \int dx = x \end{array}$$

Then

$$\int (\log{x})^2 dx = x (\log{x})^2 - 2 \int \log{x} dx$$

We use the interation by parts one more time. Set

$$\begin{array}{ll} u = log{x}, & dv = dx\\ du = \frac{1}{x}dx, & v = \int dv = \int dx = x \end{array}$$

Then

$$\int \log{x} dx = x \log{x} - \int dx = x \log{x} - x + c$$

Therefore, $${\int (\log{x})^2 dx = x(\log{x})^2 - 2(x\log{x} - x) + c}$$

(d)

$t = x + 5$ とおくと $x = t-5$ ，$dt = dx$ よって

$$\int x(x+5)^{15} dx = \int (t - 5)t^{14} dt = \frac{1}{16}(x+5)^{16} - \frac{1}{3}(x+5)^{15} + c$$

(e) Set

$$\begin{array}{ll} u = x, & dv = \cos{x} dx\\ du = dx, & v = \int dv = \int \cos{x} dx = \sin{x} \end{array}$$

Then

$$\int x \cos{x}dx = x \sin{x} - \int \sin{x} dx = x \sin{x} + \cos{x} +c$$

(f)

Set $I = \int e^{x} \sin{x}dx$ and use the integration by parts. Then

$$\begin{array}{ll} u = e^{x}, & dv = \sin{x} dx\\ du = e^{x}dx, & v = \int dv = \int \sin{x} dx = -\cos{x} \end{array}$$

Thus,

$${I = \int e^{x} \sin{x} dx = -e^{x}\cos{x} + \int e^{x} \cos{x}dx }$$ We use the integration by parts one more time. Set

$$\begin{array}{ll} u = e^{x}, & dv = \cos{x} dx\\ du = e^{x}dx, & v = \int dv = \int \cos{x} dx = \sin{x} \end{array}$$

Then

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

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

Thereforem $I$ is given by

$${I = -\frac{1}{2}[ {e^x}\,\left( \cos (x) - \sin (x) \right)]}$$

(g)Set

$$\begin{array}{ll} u = \log{(1+x^2)}, & dv = dx\\ du = \frac{2x}{1 + x^2}dx, & v = \int dv = \int dx = x \end{array}$$

Then

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

Note that

$$\int \frac{x^2}{1+x^2}dx = \int \frac{1 + x^2 - 1}{1 + x^2} dx = \int (1 - \frac{1}{1 + x^2} dx = x - \tan^{-1}{x}$$

Then

$$\int \log{(1 + x^2)}dx = x \log{(1+x^2)} - 2(x - \tan^{-1}{x}) + c$$

(h) Set

$$\begin{array}{ll} u = \tan^{-1}{x}, & dv = x dx\\ du = \frac... ...x^2}dx, & v = \int dv = \int x dx = \frac{x^2}{2} \end{array}$$

Then

$$\int \log{(1 + x^2)}dx = \frac{x^2 \tan^{-1}{x}}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2}dx$$

Note that

$$\int \frac{x^2}{1+x^2}dx = \int \frac{1 + x^2 - 1}{1 + x^2} dx = \int (1 - \frac{1}{1 + x^2} dx = x - \tan^{-1}{x}$$

Then

$$\int x \tan^{-1}{x}dx = \frac{x^2 \tan^{-1}{x}}{2} - \frac{1}{2}(x - \tan^{-1}{x}) + c$$

(i) Set

$$\begin{array}{ll} u = \log{x}, & dv = x^n dx\\ du = \frac{1}... ...& v = \int dv = \int x^n dx = \frac{x^{n+1}}{n+1} \end{array}$$

Then

$$\int x^{n} \log{x}dx = \frac{x^{n+1} \log{x}}{n+1} - \frac{1}{n+1} \int x^{n} dx$$

$$= \frac{x^{n+1} \log{x}}{n+1} - \frac{x^{n+1}}{n+1} + c$$

(j) Set

$$\begin{array}{ll} u = x^{3}, & dv = \sin{x} dx\\ du = 3x^2 dx, & v = \int dv = \int \sin{x} dx = - \cos{x} \end{array}$$

Then

$$\int x^{3} \sin{x}dx = -x^3 \cos{x} + 3 \int x^{2}\cos{x} dx$$

Similarly, $\int x^{2}\cos{x} dx$. Set

$$\begin{array}{ll} u = x^{2}, & dv = \cos{x} dx\\ du = 2x dx, & v = \int dv = \int \cos{x} dx = \sin{x} \end{array}$$

Then

$$\int x^{2} \sin{x}dx = x^2 \sin{x} + 2 \int x\sin{x} dx$$

Similarly, $\int x \sin{x} dx$. Set

$$\begin{array}{ll} u = x, & dv = \sin{x} dxとおくと\\ du = dx, & v = \int dv = \int \sin{x} dx = -\cos{x} \end{array}$$

Then

$$\int x \sin{x}dx = -x \cos{x} + \int \cos{x} dx = -x \cos{x} + \sin{x} + c$$

Therefore,

$$\int x^{3} \sin{x}dx = 6\,x\,\cos (x) - {x^3}\,\cos (x) - 6\,\sin (x) + 3\,{x^2}\,\sin (x)$$

(k) To find $\int{x \sinh{x}}dx$, one way to do is $\sinh{x} = \frac{e^{x} - e^{-x}}{2}$ or $(\sinh{x})' = \cosh{x}$．Set

$$\begin{array}{ll} u = x, & dv = \sinh{x} dx\\ du = dx, & v = \int dv = \int \sinh{x} dx = \cosh{x} \end{array}$$

Then

$$\int x \sin{x}dx = x \cosh{x} - \int \cosh{x} dx = x \cosh{x} - \sinh{x} + c$$