치환적분법

$-\frac{1}{3({x}^{3}-3)}+C$

$u=\cos \theta ,du=-\sin \theta d\theta ,{\int }_{1/\sqrt{2}}^{1}{u}^{-4}du=\frac{1}{3}\left(2\sqrt{2}-1\right)$

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 $x=3$,

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

We use the substitution $u=4-x^2$, for which $\mathrm{d}u=-2x\,\mathrm{d}x$ :

$$\begin{aligned} \int \frac{x^2-4}{\sqrt{4-x^2}}\,x\,\mathrm{d}x &=\int \frac{1}{2}\cdot\frac{ 4-x^2}{\sqrt{ 4-x^2}} ({-}2x)\,\mathrm{d}x \\ &=\frac{1}{2} \int \frac{{red}u}{\sqrt{{red}u}}\, \mathrm{d}u \\ &=\frac{1}{2}\int \sqrt{u}\,\mathrm{d}u \\ &=\frac{1}{2}\frac{u^{3/2}}{3/2}+C\\ &=\frac{1}{3}(4-x^2)^{3/2}+C \end{aligned}$$

• Solution 1: If we let $u=\sqrt{\log x}$, then $\mathrm{d}u=\dfrac{1}{2x\sqrt{\log x}}\ \mathrm{d}x$, and:

$$\begin{aligned} \int \dfrac{e^{ \sqrt{\log x}}}{ 2x\sqrt{\log x}}\ \mathrm{d}x&=\int e^{{red}u}\ \mathrm{d}u=e^u+C=e^{\sqrt{\log x}}+C \end{aligned}$$

• Solution 2: In Solution 1, we made a pretty slick choice. We might have tried to work with something a little less convenient. For example, it's not unnatural to think that $u=\log x$, $\mathrm{d}u=\dfrac{1}{x}\ \mathrm{d}x$ would be a good choice. In that case:

$$\begin{aligned} \int \dfrac{e^{\sqrt{ \log x}}}{2 x\sqrt{ \log x}}\ \mathrm{d}x&= \int \frac{e^{\sqrt{{red}u}}}{2\sqrt{{red}u}} \mathrm{d}u \\ \text{Now, we should be able to see that $ w=\sqrt{u}$, $ \mathrm{d}w = \dfrac{1}{2\sqrt{u}}\ \mathrm{d}u$ is a good choice:} \\ \int\frac{e^{ \sqrt{u}}}{ 2\sqrt{u}}\ \mathrm{d}u &=\int e^{{orange}w}\ \mathrm{d}w\\ &=e^{\sqrt{u}}+C\\ &=e^{\sqrt{\log x}}+C \end{aligned}$$