치환적분법

$3 \sqrt{x^2-2 x-6}+C$

$\sqrt{x^2-6 x+8}+C$

$-\ln 2$

$352/15$

$2/3$

$1/5$

$(1-e)/2$

$\pi/2$

$\pi/2$

$\pi/6$

The reasoning is not sound: when we do a substitution, we need to take care of the differential ($\mathrm{d}x$). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here's the way to do it:

Problem: Evaluate $\displaystyle\int (2x+1)^2 \mathrm{d}x$.

Work: We use the substitution $u=2x+1$. Then $\mathrm{d}u=2\mathrm{d}x$, so $\mathrm{d}x = \frac{1}{2}\mathrm{d}u$:

$$\begin{aligned} \int (2x+1)^2 \mathrm{d}x&=\int u^2\cdot \frac{1}{2}\ \mathrm{d}u\\ &=\frac{1}{6}u^3+C\\ &=\frac{1}{6}\left(2x+1\right)^3+C \end{aligned}$$

Perhaps shorter ways exist, but the reasoning here is valid.

Problem: Evaluate $\displaystyle\int_{0}^{\pi/4} x\tan (x^2) \ \mathrm{d}x$.

Work: We begin with the substitution $u=x^2$, $\mathrm{d}u = 2x\mathrm{d}x$:

If $u=x^2$, then $\frac{\mathrm{d}u}{\mathrm{d}x} = 2x$, so indeed $\mathrm{d}u=2x\mathrm{d}x$.

$$\begin{aligned} \int_{0}^{\pi/4} x\tan (x^2) \ \mathrm{d}x&= \int_{0}^{\pi/4} \frac{1}{2}\tan(x^2)\cdot 2x\mathrm{d}x&\text{algebra}\\ &=\int_{0}^{\pi^2/16} \frac{1}{2}\tan u\ \mathrm{d}u \\ \text{\text{Every piece is changed from $x$ to $u$: integrand, differential, limits.}} \\ &=\frac{1}{2}\int_{0}^{\pi^2/16} \dfrac{\sin u}{\cos u}\mathrm{d}u & \tan u = \frac{\sin u}{\cos u}\\ \\ \text{Now we use the substitution $v=\cos u$, $\mathrm{d}v=-\sin u \ \mathrm{d}u$:} \\ &=\frac{1}{2}\int_{\cos 0}^{\cos(\pi^2/16)} -\dfrac{1}{v}\mathrm{d}v \\ \text{\text{Every piece is changed from $u$ to $v$: integrand, differential, limits.}} \\ &=-\frac{1}{2}\int_{1}^{\cos(\pi^2/16)} \dfrac{1}{v}\mathrm{d}v& \cos(0)=1\\ &=-\frac{1}{2}\bigg[\log|v|\bigg]_{1}^{\cos(\pi^2/16)}&\text{FTC Part 2}\\ &=-\frac{1}{2}\left(\log\left(\cos(\pi^2/16)\right)-\log(1)\right)\\ &=-\frac{1}{2}\log\left(\cos(\pi^2/16)\right)& \log(1)=0 \end{aligned}$$

Using the chain rule, we see that $$\frac{\mathrm{d}}{\mathrm{d}x}\{f(g(x))\}=f'(g(x))g'(x)$$ So, $f(g(x))$ is an antiderivative of $f'(g(x))g'(x)$. All antiderivatives of $f'(g(x))g'(x)$ differ by only a constant, so:

$$\begin{aligned} \int f'(g(x))g'(x)\ \mathrm{d}x - f(g(x))&= f(g(x))+C-f(g(x))\\ &=C \end{aligned}$$

That is, our expression simplifies to some constant $C$.

Remark: since $$\int f'(g(x))g'(x)\ \mathrm{d}t - f(g(x))=C$$ we conclude $$\int f'(g(x))g'(x)\ \mathrm{d}t = f(g(x))+C$$ which is precisely how we perform substitution on integrals.

We write $u(x) = e^{x^2}$ and find $\mathrm{d}u = u'(x)\,\mathrm{d}x=2x e^{x^2}\mathrm{d}x$. Note that $u(1)=e^{1^2}=e$ when $x=1$, and $u(0)=e^{0^2}=1$ when $x=0$. Therefore:

$$\begin{aligned} \int_{0}^{1} x e^{x^2} \cos ( e^{x^2}) \, \mathrm{d}x &= \frac{1}{2}\int_{x=0}^{x=1} \cos ( u(x)) u'(x)\,\mathrm{d}x \\ &= \frac{1}{2}\int_{u=1}^{u=e} \cos({red}u)\,\mathrm{d}u\\ &= \frac{1}{2}\bigg[\sin(u) \bigg]_1^e = \frac{1}{2}\big( \sin(e) - \sin(1) \big). \end{aligned}$$

Substituting $y=x^3$, $\mathrm{d}y=3x^2\ \mathrm{d}x$ :

$$\begin{aligned} \int_1^2 x^2 f( x^3)\, \mathrm{d}x = \frac{1}{3}\int_1^8 f( y)\, \mathrm{d}y =\frac{1}{3} \end{aligned}$$

Setting $u=x^3+1$, we have $\mathrm{d}u = 3x^2\,\mathrm{d}x$ and so

$$\begin{aligned} \int \frac{ x^2\,\mathrm{d}x}{{( x^3+1)}^{101}} &= \int \frac{ \mathrm{d}u/3}{ u^{101}}\\ &=\frac{1}{3}\int u^{-101}\ \mathrm{d}u\\ &=\frac{1}{3}\cdot\dfrac{u^{-100}}{-100}\\ &= -\frac{1}{3\times 100 u^{100}} + C\\ &=-\frac{1}{300{(x^3+1)}^{100}} + C \end{aligned}$$

Setting $u=\log x$, we have $\mathrm{d}u = \frac{1}{x}\,\mathrm{d}x$ and so

$$\int_{e}^{e^4} \frac{ \mathrm{d}x}{{blue}x\cdot \log x} = \int_{x=e}^{x=e^4} \frac1{ \log x} \cdot \frac{1}{x}\,\mathrm{d}x = \int_{u=1}^{u=4} \frac{1}{{red}u}\, \mathrm{d}u,$$

since $u=\log(e)=1$ when $x=e$ and $u=\log(e^4)=4$ when $x=e^4$. Then, by the Fundamental Theorem of Calculus Part 2,

$$\int_{1}^{4} \frac{1}{u}\, \mathrm{d}u = \Big[\log |u| \Big]_{1}^{4} = \log 4 - \log 1 = \log 4.$$

Setting $u=1+\sin x$, we have $\mathrm{d}u = \cos x~\mathrm{d}x$ and so

$$\int_{0}^{\pi/2} \frac{ \cos x} { 1+\sin x} \mathrm{d}x = \int_{x=0}^{x=\pi/2} \frac{1}{ 1+\sin x}\, \cos x ~\mathrm{d}x = \int_{u=1}^{u=2} \frac{ \mathrm{d}u}{{red}u}$$

since $u=1+\sin 0=1$ when $x=0$ and $u=1+\sin(\pi/2)=2$ when $x=\pi/2$. Then, by the Fundamental Theorem of Calculus Part 2,

$$\int_{u=1}^{u=2} \frac{\mathrm{d}u}{u} = \Big[\log|u| \Big]_{1}^{2} = \log 2$$

Setting $u=\sin x$, we have $\mathrm{d}u = \cos x~\mathrm{d}x$ and so

$$\int_{0}^{\pi/2} \cos x \cdot (1+ \sin^2 {red}x) \mathrm{d}x = \int_{x=0}^{x=\pi/2} (1+ \sin^2 x)\cdot \cos x ~\mathrm{d}x = \int_{u=0}^{u=1} (1+ u^2) \, \mathrm{d}u,$$

since $u=\sin 0=0$ when $x=0$ and $u=\sin(\pi/2)=1$ when $x=\pi/2$. Then, by the Fundamental Theorem of Calculus Part 2,

$$\int_{0}^{1} (1+u^2) \,\mathrm{d}u = \left[u+\frac{u^3}{3} \right]_{0}^{1} =\left(1+\frac{1}{3}\right) -0 = \frac{4}{3}.$$

Substituting $t=x^2-x$, $\mathrm{d}t = (2x-1)\,\mathrm{d}x$ and noting that $t=0$ when $x=1$ and $t=6$ when $x=3$,

$$\begin{aligned} \int_1^3 (2x-1)e^{ x^2-x} \mathrm{d}x &= \int_0^6 e^{{red}t}\ \mathrm{d}t =\big[e^t\big]_0^6 =e^6-1 \end{aligned}$$