Integration by Parts

calculus

Lecture 5.1 - Integrals Involving Inverse Trig Functions

https://youtu.be/yeaI1XhQWVQ

Review of inverse functions

f(x) and f^-1(x) are inverse functions, so $f(f^{-1}(x))=x$

Using the chain rule, also, $f^{\prime }(f^{-1}(x))\cdot (f^{-1}{)}^{\prime }(x)=1$

..which would mean: $(f^{-1}{)}^{\prime }(x)=\frac{1}{f^{\prime }(f^{-1}(x))}$

In the last class, they showed how $\frac{d}{dx}[e^x]=e^x$, therefore $\frac{d}{dx}[ln(x)]=\frac{1}{x}$.

$$\begin{array}{r}f(x)=e^x\\ \\ f^{\prime }(x)=e^x\\ \\ f^{-1}(x)=ln(x)\\ \\ \frac{d}{dx}[ln(x)]=\frac{1}{e^{ln(x)}}=\frac{1}{x}\end{array}$$

Inverse trig functions

$\frac{d}{dx}[si{n}^{-1}(x)]=\frac{1}{\sqrt{1-x^2}}$

$$\begin{array}{r}f(x)=sin(x)\\ \\ f^{\prime }(x)=cos(x)\\ \\ f^{-1}(x)=si{n}^{-1}(x)\\ \\ f^{\prime -1}(x)=\frac{1}{f^{\prime }(f^{-1}(x))}\\ \\ \frac{d}{dx}[si{n}^{-1}(x)]=\frac{1}{cos(si{n}^{-1}(x))}\end{array}$$

$\frac{d}{dx}[ta{n}^{-1}(x)]=\frac{1}{\sqrt{x^2+1}}$

$$\begin{array}{r}f(x)=tan(x)\\ \\ f^{\prime }(x)=se{c}^2(x)\\ \\ se{c}^2(x)=\frac{1}{co{s}^2(x)}\\ \\ f^{-1}(x)=ta{n}^{-1}(x)\\ \\ (f^{-1}{)}^{\prime }(x)=\frac{1}{f^{\prime }(f^{-1}(x))}\\ \\ \frac{d}{dx}[ta{n}^{-1}(x)]=\frac{1}{se{c}^2(ta{n}^{-1}(x))}\\ \\ =co{s}^2(ta{n}^{-1}(x))\end{array}$$

Results

I feel like his diagram is really useful here as well b/c it shows how you do (cos(tan^-1(x))) by drawing out the triangle:

so b/c of Fundamental theorem of calculus,

$$\begin{array}{r}\int \frac{1}{\sqrt{1-x^2}}dx=si{n}^{-1}(x)+C\\ \\ \int \frac{1}{x^2+1}dx=ta{n}^{-1}(x)+C\end{array}$$

Combined with u-substitution

To make these work, you need to remember how to find perfect square trinomials.

$\int \frac{1}{x^2+4x+8}dx$

$$\begin{array}{rlr}& \int \frac{1}{x^2+4x+8}dx& \\ & & \\ & \int \frac{1}{(x^2+4x+4)+4}dx& \text{"borrow 4 from 8 to make ’perfect square’}\\ & & \\ & \int \frac{1}{(x+2{)}^2+4}dx& \\ & & \\ & \frac{1}{4}\int \frac{1}{\frac{(x+2{)}^2}{4}+1}dx& \\ & & \\ & \frac{1}{4}\int \frac{1}{(\frac{x+2}{2}{)}^2+1}dx& \\ & & \\ u& =\frac{x+2}{2}& \\ & & \\ du& =\frac{1}{2}dx& \\ & & \\ & \frac{1}{2}\int \frac{1}{u^2+1}du& \text{move 1/2 from the outside into the du}\\ & & \\ & \frac{1}{2}ta{n}^{-1}(u)+C& \\ & & \\ & \frac{1}{2}ta{n}^{-1}(\frac{x+2}{2})+C& \end{array}$$

$\int \frac{1}{\sqrt{4-2x-x^2}}dx$

$$\begin{array}{rl}& \int \frac{1}{\sqrt{4-2x-x^2}}dx\\ & \\ & \int \frac{1}{\sqrt{5-(x^2+2x+1)}}dx\\ & \\ & \int \frac{1}{\sqrt{5-(x+1{)}^2}}dx\\ & \\ & \int \frac{1}{\sqrt{5(1-\frac{(x+1{)}^2}{5})}}dx\\ & \\ & \int \frac{1}{\sqrt{5}\sqrt{1-(\frac{x+1}{\sqrt{5}}{)}^2}}dx\\ & \\ u& =\frac{x+1}{\sqrt{5}}\\ & \\ du& =\frac{1}{\sqrt{5}}dx\\ & \\ & \int \frac{1}{\sqrt{1-u^2}}du\\ & \\ & si{n}^{-1}(u)+C\\ & \\ & si{n}^{-1}(\frac{x+1}{\sqrt{5}})+C\end{array}$$

Homework

6.1 43-52, 61-78

47 $\int \frac{5}{\sqrt{x^4-16x^2}}dx$

$$\begin{array}{r}\int \frac{5}{\sqrt{x^4-16x^2}}dx\end{array}$$

I need to remove x^2 from the denominator there. Not entirely sure how to do it in a way that balances out.

50

$$\begin{array}{r}\int \frac{2}{\sqrt{-x^2+6x+7}}dx\end{array}$$

53 validate

$$\begin{array}{rlr}& \int \frac{x^2}{(x^3+3{)}^2}dx& \\ & & \\ u& =x^3+3& \\ & & \\ du& =3x^{2}dx& \\ & & \\ & \int \frac{\frac{1}{3}du}{u^2}& \\ & & \\ & \frac{1}{3}\int \frac{du}{u^2}& \\ & & \\ & \frac{1}{3}3u^3& \text{1/3rd of inverse u2}\\ & & \\ & u^3& \end{array}$$

66

$$\begin{array}{r}\int \frac{2}{4x^2+1}dx\end{array}$$

72

$$\begin{array}{r}\int \frac{x^3}{x^2+9}dx\end{array}$$

74

$$\begin{array}{r}\int \frac{sin(x)}{co{s}^2(x)+1}dx\end{array}$$

6.2 5-49

5 $\int x\sin xdx$

$$\begin{array}{rl}& \int x\sin xdx\\ & \\ u& =x\\ & \\ du& =dx\\ & \\ v& =cos(x)\\ & \\ dv& =sin(x)dx\\ & \\ & =uv-\int vdu\\ & \\ & =x\cdot cos(x)-\int cos(x)dx\\ & \\ & =x\cdot cos(x)-sin(x)\end{array}$$

10 $\int x^3{e}^{x}dx$

$$\begin{array}{rl}\int x^3{e}^{x}dx& \\ & \\ u& =x^3\\ & \\ du& =3x^2\\ & \\ v& =e^x\\ & \\ dv& =e^x\\ & \\ x^3{e}^x-\int e^{x}3x^2& \\ & \\ x^3{e}^x-e^{x}3x^2& \end{array}$$

17 $\int si{n}^{-1}(x)dx$

30 $\int x\sqrt{x-2}dx$

$$\begin{array}{rl}& \text{6.2, number 30}\\ & \\ & \int x\sqrt{x-2}dx\\ & \\ u& =x\\ & \\ du& =dx\\ & \\ v& =\frac{2}{3}(x-2{)}^{\frac{3}{2}}\\ & \\ dv& =(x-2{)}^{.5}dx\\ & \\ & =x\cdot \frac{2}{3}(x-2{)}^{\frac{3}{2}}-\int \frac{2}{3}(x-2{)}^{\frac{3}{2}}dx\\ & \\ & =x\cdot \frac{2}{3}(x-2{)}^{\frac{3}{2}}-(x-2{)}^{.5}+C\end{array}$$

36 $\int e^{2x}cos(e^x)dx$

39 $\int e^{\sqrt{x}}dx$

42 ${\int }_{-1}^{1}x{e}^{-x}dx$

46 ${\int }_0^1{x}^3{e}^{x}dx$

Quizes

Quiz 1

1

${\int }_1^{4}5x{e}^{x^2}dx$ turns into ${\int }_A^{B}C{e}^{u}du$. What are A B & C?

2

value of ${\int }_0^{0.6}\frac{1}{x^2+4}dx$?

$$\begin{array}{r}{\int }_0^{0.6}\frac{1}{x^2+4}dx\\ \\ \frac{1}{4}{\int }_0^{0.6}\frac{1}{(\frac{x}{2}{)}^2+1}dx\\ \\ u=\frac{x}{2}du=\frac{1}{2}dx\frac{1}{4}{\tan }^{-1}(u){|}_0^{0.6}\\ \\ \frac{1}{4}({\tan }^{-1}(0.6)-{\tan }^{-1}(0))\end{array}$$

u = (1/2 x) du = 1/2 dx 1/4 tan-1(x) 1/4 tan-1(x/2)

3

No clue. guessing.

$$\begin{array}{r}\int si{n}^3(x)co{s}^2(x)dx\\ \\ u=cos(x)\end{array}$$

4

$$\begin{array}{r}(u+2)(u+2)=u^{2}4u2u+4x+8+3\end{array}$$

5

$$\begin{array}{r}{\int }_0^{\frac{\pi }{3}}xcos(2x)dxu=xdu=dxv=sin(2x)\cdot 2dv=cos(2x)dxu\cdot v-\int vdu2x\cdot sin(2x)-\int sin(2x)\cdot 2dx2x\cdot sin(2x)+cos(2x)\end{array}$$

Quiz 2

1

$$\begin{array}{rl}& {\int }_1^{4}3\sqrt{x}{e}^{\sqrt{x}}dx\\ & \\ u& =\sqrt{x}\\ & \\ du& =\frac{2}{3}{x}^{\frac{3}{2}}dx\\ & \\ & {\int }_1^{4}3u{e}^{u}dx\\ & \\ & \frac{3}{2}{u}^2{e}^u+C{|}_1^4\end{array}$$

2

exact value of:

$$\begin{array}{rl}{\int }_0^{0.6}\frac{1}{x^2+4}dx& \\ & \\ 4{\int }_0^{0.6}\frac{1}{(\frac{x}{2}{)}^2+1}dx& \\ & \\ u& =\frac{x}{2}\\ & \\ 4ta{n}^{-1}(u){|}_0^{0.6}& \\ & \\ 4ta{n}^{-1}(0.6/2)& \end{array}$$

So I don’t pull the 4 out. I think it’s 1/4

4

$$\begin{array}{rl}\int \frac{x^2+4x+3}{x-2}dx& \\ & \\ u& =x-2\\ & \\ du& =\frac{1}{2}{x}^2\\ & \\ \int \frac{2u+(4u+4\cdot 2)+3}{u}& \\ & \\ \int \frac{6u+15}{u}& \\ & \\ \int 6u+15\cdot \frac{1}{u}& \\ & \\ & =3u^2\cdot \ln |u|+C\\ & \\ & =3(x-2{)}^2\cdot \ln |x-2|+C\end{array}$$

Quiz 3/4

1.

Consider the integral ${\int }_1^{2}4x^{3}cos(x^2)dx$ w/ $u=x^2$, it becomes ${\int }_A^{B}C{u}^{D}cos(u)du$. What is a/b/c/d?

a=1 b=4 (b/c we need to account for previous squaring C = 4 (constant) D = 3/2 (1 + 1/2 of x^2)

—

Seems like this wasn’t quite right. Instead, when u=x^2, du = 2x, which means the 4x^3 would be udu. This means the C constant would be 2, not 4. This also means that D would be 1, not 1.5.

2

Find the exact value

$$\begin{array}{rl}{\int }_0^{0.6}\frac{1}{x^2+4}dx& \\ & \\ \frac{1}{4}{\int }_0^{0.6}\frac{1}{\frac{x^2}{4}+1}dx& \\ & \\ \frac{1}{4}{\int }_0^{0.6}\frac{1}{(\frac{x}{2}{)}^2+1}dx& \\ & \\ u& =\frac{x}{2}\\ & \\ du& =2dx\\ & \\ \frac{1}{2}{\int }_0^{0.3}\frac{1}{(u{)}^2+1}du& \\ & \\ \frac{1}{2}ta{n}^{-1}(u){|}_0^{0.6}& \\ & \\ \frac{1}{2}ta{n}^{-1}(\frac{x}{2}){|}_0^{0.6}& \\ & \\ \frac{1}{2}ta{n}^{-1}(0.3)& \end{array}$$

I think going to 0.3 is because x went from x/2 to just u, so divide the B value.

3

Find $\int si{n}^5(x)dx$ where $u=cos(x)$.

$$\begin{array}{r}\int si{n}^5(x)dx\\ \\ \int (1-co{s}^2(x){)}^2\cdot sin(x)dx\\ \\ u=cos(x)\\ \\ du=sin(x)\\ \\ \int (1-u^2{)}^{2}du\end{array}$$

u = cos(x) means du = -sin(x), so the final value should be $-\int (1-u^2{)}^{2}du$.

4

$$\begin{array}{r}\int \frac{x^2+4x+3}{x-2}dx\\ \\ u=x-2\\ \\ du=dx\\ \\ \int \frac{x^2+4x+3}{u}du\end{array}$$

… unsure?

Need to do division by polynomials

5

$$\begin{array}{rl}& {\int }_0^{\frac{\pi }{3}}xcos(2x)dx\\ & \\ u& =x\\ & \\ du& =dx\\ & \\ v& =\frac{1}{2}sin(2x)\\ & \\ dv& =cos(2x)dx\\ & \\ & \frac{x}{2}sin(2x){|}_0^{\frac{\pi }{3}}-{\int }_0^{\frac{\pi }{3}}\frac{1}{2}sin(2x)dx\\ & \\ & \frac{x}{2}sin(2x){|}_0^{\frac{\pi }{3}}-cos(2x){|}_0^{\frac{\pi }{3}}\\ & \\ & (\frac{\pi }{6}sin(2\pi )-sin(\frac{\pi }{3}))-(cos(\frac{2\pi }{3})-cos(0))\\ & \end{array}$$

Next to last line should have been

$$\begin{array}{rl}& \frac{x}{2}sin(2x){|}_0^{\frac{\pi }{3}}+\frac{1}{4}cos(2x){|}_0^{\frac{\pi }{3}}\\ & \\ & (\frac{\pi }{6}sin(2\pi )-sin(\frac{\pi }{3}))-(\frac{1}{4}cos(\frac{2\pi }{3})+\frac{1}{4}cos(0))\\ & \end{array}$$