부분적분법

We integrate by parts, using $u = \arctan(2y)$, $\mathrm{d}v = 4y \,\mathrm{d}y$, so that $v=2y^2$ and $\mathrm{d}u= \frac{2\, \mathrm{d}y}{1+(2y)^2}$:

$$\begin{aligned} \int 4y\arctan(2y) \,\mathrm{d}y = \underbrace{2y^2\arctan(2y)}_{uv} - \int\underbrace{ \frac{4y^2}{(2y)^2+1} \,\mathrm{d}y}_{v\mathrm{d}u} \end{aligned}$$

The integrand $\frac{4y^2}{(2y)^2+1}$ is a rational function. So the remaining integral can be evaluated using the method of partial fractions, starting with long division. But it is easier to just notice that $\frac{4y^2}{4y^2+1} = \frac{4y^2+1}{4y^2+1} - \frac{1}{4y^2+1}$. We therefore have:

$$\begin{aligned} \int \frac{4y^2}{4y^2+1} \,\mathrm{d}y = \int \left(1 - \frac{1}{4y^2+1}\right)\,\mathrm{d}y = y - \frac12\arctan(2y) + C \end{aligned}$$

The final answer is then

$$\begin{aligned} \int 4y\arctan(2y) \,\mathrm{d}y = 2y^2\arctan(2y) - y + \frac{1}{2}\arctan(2y) + C \end{aligned}$$

It's not obvious where to start, but in general it's nice to have the arguments of our trig functions the same. So, we use the identity $\sin(2x)=2\sin x \cos x$.

$$\begin{aligned} \int e^{\cos x}\sin(2x)\mathrm{d}x&=2\int e^{\cos x}\cos x \sin x\,\mathrm{d}x \\ \text{Now we can use the substitution $w=\cos x$, $\mathrm{d}w=-\sin x \mathrm{d}x$.} \\ &=-2\int we^w\mathrm{d}w \\ \text{From here the integral should look more familiar. We can use integration by parts with $u=w$, $\mathrm{d}v = e^w\mathrm{d}w$, $\mathrm{d}u=\mathrm{d}w$, and $v=e^w$.} \\ &=-2\left[\underbrace{we^w}_{uv} - \int \underbrace{e^w\mathrm{d}w}_{v\mathrm{d}u}\right]\\ &=2e^w\left[1-w\right]+C\\ &=2e^{\cos x}[1-\cos x]+C \end{aligned}$$

We've got an integrand that consists of several functions multiplied together, and no obvious substitution. So, we think about integration by parts. We know an antiderivative for $\frac{1}{(1-x)^2}$, because we know $\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{1-x}=\frac{1}{(1-x)^2}$. So let's try $\mathrm{d}v = \frac{\mathrm{d}x}{(1-x)^2}$ and $u=xe^{-x}$. Then $v=\frac{1}{1-x}$ and $\mathrm{d}u = (1-x)e^{-x}\,\mathrm{d}x$. So, by integration by parts,

$$\begin{aligned} \int\underbrace{xe^{-x}}_{u}\underbrace{ \frac{\mathrm{d}x}{(1-x)^2}}_{\mathrm{d}v} &= \underbrace{\frac{xe^{-x}}{1-x}}_{uv} -\int\underbrace{\frac{1}{1-x}}_{v}\underbrace{(1-x)e^{-x}\,\mathrm{d}x}_{\mathrm{d}u} \\ &=\frac{xe^{-x}}{1-x} -\int e^{-x}\,\mathrm{d}x \\ &=\frac{xe^{-x}}{1-x} + e^{-x} + C =\frac{e^{-x}}{1-x} + C \end{aligned}$$

(a) The “parts" in the integrand are powers of sine. Looking at the right hand side of the reduction formula, we see that it looks a little like the derivative of $\sin^{n-1}x$, although not exactly. So, let's integrate by parts with $u=\sin^{n-1}x$ and $\mathrm{d}v=\sin x\,\mathrm{d}x$, so that $\mathrm{d}u=(n-1)\sin^{n-2}x\cos x$ and $v=-\cos x$.

$$\begin{aligned} \int\sin^nx\,\mathrm{d}x &=\underbrace{-\sin^{n-1}x\ \cos x}_{uv}+\underbrace{(n-1)\int \cos^2x\ \sin^{n-2}x\ \mathrm{d}x}_{-\int v\mathrm{d}u} \\ \text{Using the identity $\sin^2 x + \cos^2 x =1 $,} \\ &=-\sin^{n-1}x\ \cos x+(n-1)\int (1-\sin^2 x)\sin^{n-2}x\ \mathrm{d}x\\ &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \mathrm{d}x -(n-1)\int\sin^{n}x\ \mathrm{d}x \end{aligned}$$

Moving the last term on the right hand side to the left hand side gives

$$\begin{aligned} n\int\sin^nx\,\mathrm{d}x &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \mathrm{d}x \end{aligned}$$

Dividing across by $n$ gives the desired reduction formula.

(b) By the reduction formula of part (a), if $n \ge 2$,

$$\begin{aligned} \int_0^{\pi/2}\sin^n(x)\,\mathrm{d}x= \frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}(x)\,\mathrm{d}x \end{aligned}$$

since $\sin 0=\cos\frac{\pi}{2}=0$. Applying this reduction formula, with $n=8,6,4,2$:

$$\begin{aligned} \int_0^{\pi/2}\sin^8(x)\,\mathrm{d}x &=\frac{7}{8}\int_0^{\pi/2}\sin^6(x)\,\mathrm{d}x =\frac{7}{8}\cdot\frac{5}{6}\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\int_0^{\pi/2}\sin^2(x)\,\mathrm{d}x \\ &=\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\int_0^{\pi/2}\,\mathrm{d}x =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} =\frac{35}{256}\pi \end{aligned}$$

Using a calculator, we see this is approximately $0.4295$.

For a fixed value of $x$, if we rotate about the $x$-axis, we form a washer of inner radius $B(x)$ and outer radius $T(x)$ and hence of area $\pi [T(x)^2 - B(x)^2]$. We integrate this function from $x=0$ to $x=3$ to find the total volume $V$:

$$\begin{aligned} V &= \int_0^3 \pi [T(x)^2 - B(x)^2]\,\mathrm{d}x \\ &= \pi \int_0^3 (\sqrt{x}e^{3x})^2 - (\sqrt{x}(1+2x))^2 \,\mathrm{d}x \\ &= \pi \int_0^3 \big( xe^{6x} - (x+4x^2+4x^3) \big) \,\mathrm{d}x \\ &= \pi \int_0^3 xe^{6x} \,\mathrm{d}x - \pi \Big[ \frac{x^2}{2} + \frac{4x^3}{3} + x^4\Big]_{0}^3 \\ &= \pi \int_0^3 xe^{6x} \,\mathrm{d}x - \pi \Big[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4\Big] \end{aligned}$$

For the first integral, we use integration by parts with $u(x) = x$, $\mathrm{d}v = e^{6x}\mathrm{d}x$, so that $\mathrm{d}u=\mathrm{d}x$ and $v(x)=\frac16e^{6x}$:

$$\begin{aligned} \int_0^3 xe^{6x} \,\mathrm{d}x &=\underbrace{ \frac{xe^{6x}}{6}}_{uv}\bigg|_{0}^3 - \int_0^3 \underbrace{\frac{1}{6}e^{6x} \,\mathrm{d}x }_{v\mathrm{d}u}\\ &= \frac{3e^{18}}{6} - 0 - \frac{1}{36} e^{6x} \bigg|_{0}^3 = \frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg). \end{aligned}$$

Therefore, the total volume is

$$\begin{aligned} V = \pi \bigg[\frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg) \bigg] - \pi \bigg[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4 \bigg] = \pi \bigg( \frac{17 e^{18}-4373}{36} \bigg). \end{aligned}$$

To get rid of the square root in the argument of $f''$, we make the change of variables (also called “substitution") $x=t^2,\ \mathrm{d}x=2t\,\mathrm{d}t$.

$$\begin{aligned} \int_0^4 f''\big(\sqrt{x}\big)\,\mathrm{d}x &= 2\int_0^2 tf''(t)\,\mathrm{d}t \end{aligned}$$

Then, to convert $f''$ into $f'$, we integrate by parts with $u=t$, $\mathrm{d}v=f''(t)\,\mathrm{d}t$, $v=f'(t)$.

$$\begin{aligned} \int_0^4 f''\big(\sqrt{x}\big)\,\mathrm{d}x &= 2\bigg\{\Big[\underbrace{tf'(t)}_{uv}\Big]_0^2-\int_0^2\!\!\!\underbrace{ f'(t)\,\mathrm{d}t}_{v\mathrm{d}u}\bigg\} \\ &=2\Big[tf'(t)-f(t)\Big]_0^2 \\ &=2\big[2f'(2)-f(2)+f(0)\big]=2\big[2\times 4-3+1\big]\\ &=12 \end{aligned}$$