u-substitution

I believe this is a mechanism of applying the Chain rule to simplify calculating differentials.

We should be able to back-test our results by integrating the result and making sure we get the original function.

Given this equation, we should be able to realize that when we differentiate $x^2-3x+5$, we get $2x-3$. Therefore:

$$\begin{array}{rl}& \int (x^2-3x+5{)}^4(2x-3)dx\\ & \\ u& =x^2-3x+5\\ & \\ \frac{du}{dx}& =2x-3\\ & \\ du& =(2x-3)dx\\ & \\ & \int u^{4}du\\ & \\ & =\frac{1}{5}{u}^5+C\\ & \\ & =\frac{1}{5}(x^2-3x+5{)}^5+C\end{array}$$

Once we get $u$, then we can do differentiation to figure out $du$.

Practice

1: $\int e^{-4x}dx$

I’m not actually sure about this b/c $e^x$ differentiates to itself, as best I know.

Okay, so the answer here is to try it anyhow. -4x is the only thing which could be in terms of x, so we try that as our U. In order to make introducing a du possible, we have to introduce a 1/4th constant, which should work itself out.

$$\begin{array}{rl}& \int e^{-4x}dx\\ & \\ u& =-4x\\ & \\ du& =-4\\ & \\ & =-\frac{1}{4}{e}^{-4x}\cdot -4dx\\ & \\ & =-\frac{1}{4}{e}^u\cdot du\\ & \\ & =-\frac{1}{4}(e^u+C)\\ & \\ & =-\frac{1}{4}{e}^{-4x}+C\end{array}$$

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

Not immediately apparent to me, so we’ll try the FAFO trick.

$$\begin{array}{r}\int \frac{2}{4x-1}dx\\ \\ u=4x\\ \\ du=4\\ \\ \int 2\cdot \frac{2}{u-1}\\ \\ \int \frac{4}{2(u-1)}\\ \\ \int \frac{du}{2(u-1)}\end{array}$$

That.. doesn’t seem particularly helpful? Teacher expanded it and made $u=4x-1$.

$$\begin{array}{rl}\int \frac{2}{4x-1}dx& \\ & \\ u& =4x-1\\ & \\ du& =4dx\\ & \\ \int \frac{2}{u}& \\ & \\ \frac{1}{2}\int 2\cdot \frac{2}{u}& \\ & \\ \frac{1}{2}\int \frac{du}{u}& \\ & \\ \frac{1}{2}\int \frac{1}{u}& \\ & \\ \frac{1}{2}\log |u|+C& \\ & \\ \frac{1}{2}\log |4x-1|+C& \end{array}$$

3: $\int \sin (x)\cos (x)dx$

$$\begin{array}{rl}& \int \sin (x)\cos (x)dx\\ & \\ u& =\sin (x)\\ & \\ \frac{du}{dx}& =\cos (x)\\ & \\ du& =\cos (x)dx\\ & \\ & \int udu\\ & \\ & =\frac{1}{2}{u}^2+C\\ & \\ & =\frac{1}{2}{\sin }^2(x)+C\end{array}$$

https://www.youtube.com/watch?v=43VM02uTHPE

1

$$\begin{array}{rl}& \int 2x\sqrt{x+5}dx\\ & \\ u& =x+5\\ & \\ du& =dx\\ & \\ x& =u-5\\ & \\ 2x& =2u-10\\ & \\ & \int (2u-10)\sqrt{u}du\\ & \\ & \int 2u^{\frac{3}{2}}-10u^{\frac{1}{2}}du\\ & \\ & =\frac{2}{\frac{5}{2}}{u}^{\frac{5}{2}}-10\cdot \frac{2}{3}{u}^{\frac{3}{2}}\\ & \\ & =\frac{4}{5}{u}^{\frac{5}{2}}-\frac{20}{3}{u}^{\frac{3}{2}}+C\\ & \\ & =\frac{4}{5}(x+5{)}^{\frac{5}{2}}-\frac{20}{3}(x+5{)}^{\frac{3}{2}}+C\\ & \\ & \text{optional, but same as..}\\ & \\ & =\frac{4}{5}(x+5{)}^2\cdot \sqrt{x+5}-\frac{20}{3}(x+5)\sqrt{x+5}+C\end{array}$$

2

$$\begin{array}{rl}& \int \tan (x)dx\\ & \\ & \int \frac{\sin (x)}{\cos (x)}dx\\ & \\ u& =\cos (x)\\ & \\ du& =-\sin (x)dx\\ & \\ & =-\int \frac{-sin(x)}{cos(x)}dx\\ & \\ & =-\int \frac{du}{u}\\ & \\ & =-\ln |u|+C\\ & \\ & =-\ln |\cos (x)|+C\\ & \\ & \text{sometimes rewritten as..}\\ & \\ & =\ln (|\cos (x){|}^{-1})+C\\ & \\ & =\ln (|\sec (x)|+C\end{array}$$

3

$$\begin{array}{r}\int \frac{x^3}{\sqrt{x^2+1}}dx\\ \\ u=x^2+1\\ \\ du=2x\end{array}$$

… not sure where to go from here?

$$\begin{array}{rl}& \int \frac{x^3}{\sqrt{x^2+1}}dx\\ & \\ u& =x^2+1\\ & \\ du& =2xdx\\ & \\ & \text{"borrow" one of the x’s from the top, and multiply by .5 to counteract 2x}\\ & \\ & \frac{1}{2}\int \frac{x^2}{\sqrt{x^2-1}}2xdx\\ & \\ & \text{Because we know that u=x2+1, x2=u−1..}\\ & \\ & \frac{1}{2}\int \frac{u-1}{\sqrt{u}}\cdot du\\ & \\ & \frac{u}{\sqrt{u}}=\sqrt{u}=u^{\frac{1}{2}}\\ & \\ & \frac{1}{\sqrt{u}}=u^{-\frac{1}{2}}\\ & \\ & \frac{1}{2}\int u^{.5}-u^{-.5}\\ & \\ & \text{now anti-differentiate}\\ & \\ \frac{1}{2}[\frac{2}{3}{u}^{\frac{3}{2}}-2u^{\frac{1}{2}}]+C& \\ & \\ & =\frac{1}{3}(x^2+1{)}^{\frac{3}{2}}-(x^2+1{)}^{\frac{1}{2}}+C\end{array}$$

4

use the trig identity: ${\tan }^2(x)={\sec }^2(x)-1$

$$\begin{array}{rl}& \int {\tan }^2(x)dx\\ & \\ & \int {\sec }^2(x)-1dx\\ & \\ & \int {\sec }^2(x)dx-\int 1dx\\ & \\ & tan(x)-x\end{array}$$

u-substitution w/ definite integrals

https://www.youtube.com/watch?v=r4Gjl7GNM7I

When dealing w/ definite integrals, when you’ve gone from $dx$ to $du$, you have to change the range of the definite integral. One mechanism of doing that is to substitute in the a and b values into the u function for x to get the new value.

1

$$\begin{array}{rl}& {\int }_1^{6}x\sqrt{x+3}\\ & \\ u& =x+3\\ & \\ du& =dx\\ & \\ & {\int }_4^9(u-3)\cdot \sqrt{u}du\\ & \\ & {\int }_4^9{u}^{\frac{3}{2}}-3u^{\frac{1}{2}}du\\ & \\ & \frac{2}{5}{u}^{\frac{5}{2}}-2u^{\frac{3}{2}}{|}_4^9\\ & \\ & =(\frac{2}{5}(9{)}^{\frac{5}{2}}-2(9{)}^{\frac{3}{2}})-(\frac{2}{5}(4{)}^{\frac{5}{2}}-2(4{)}^{\frac{3}{2}})\\ & \\ & =46.4\end{array}$$

2

$$\begin{array}{rl}{\int }_0^{\frac{\pi }{6}}& \sin (x){\cos }^2(x)dx\\ & \\ u& =cos(x)\\ & \\ du& =-sin(x)dx\\ & \\ -{\int }_0^{\frac{\pi }{6}}& -\sin (x){\cos }^2(x)dx\\ & \\ -{\int }_1^{\frac{\sqrt{3}}{2}}& u^2\cdot dudx\\ & \\ {\int }_{\frac{\sqrt{3}}{2}}^1& u^2\cdot dudx\\ & \\ & =\frac{1}{3}{u}^3{|}_{\frac{\sqrt{3}}{2}}^1\\ & \\ & =\frac{1}{3}{1}^3-\frac{1}{3}(\frac{\sqrt{3}}{2}{)}^3\\ & \\ & =.116827\end{array}$$

Integrating powers of trig functions

https://www.youtube.com/watch?v=tlKBHHGfIso

We can solve powers of trig functions w/ substitutions like $f(x)={\sin }^m(x){\cos }^n(x)$.

• ${\sin }^2(x)+{\cos }^2(x)=1$
• ${\sin }^2(x)=\frac{1-\cos (2x)}{2}$
• ${\cos }^2(x)=\frac{1+\cos (2x)}{2}$

Useful b/c they’re reducing the power of the sin/cos.

$$\begin{array}{r}\int {\sin }^2(x)dx\\ \\ \int {\sin }^2(x)=\frac{1-\cos (2x)}{2}dx\\ \\ u=2x\\ \\ du=2dx\\ \\ \frac{1}{2}\int \frac{1-cos(2x)}{2}2dx\\ \\ \frac{1}{4}\int 1-\cos (u)du\\ \\ \frac{1}{4}[u-\sin (u)]+C\\ \\ \frac{1}{4}[2x-\sin (2x)]+C\\ \\ \frac{1}{2}x-\frac{1}{4}\sin (2x)\end{array}$$

Odd powers & pythagoriean identity

Odd powers are easier w/ Pythagoriean identity.

$$\begin{array}{rlr}& \int {\cos }^5(x)dx& \\ & & \\ & \int {\cos }^4(x)\cdot \cos (x)dx& \\ & & \\ & \int [{\cos }^2(x){]}^2\cdot cos(x)dx& \\ & & \\ & \int [1-{\sin }^2(x){]}^2\cdot \cos (x)dx& \text{pythagoriean identity}\\ & & \\ u& =sin(x)& \\ & & \\ du& =cos(x)dx& \\ & & \\ & \int [1-u^2(x){]}^{2}du& \\ & & \\ & \int 1-2u^2+u^{4}du& \\ & & \\ & u-\frac{2}{3}{u}^3+\frac{1}{5}{u}^5+C& \\ & & \\ & sin(x)-\frac{2}{3}si{n}^3(x)+\frac{1}{5}{\sin }^5(x)+C& \end{array}$$

Even powers and half-angle identities

e.g. $\int si{n}^6(x)dx=\int [si{n}^2(x){]}^{3}dx$

Even powers usually require a bunch more work.

$$\begin{array}{r}\int si{n}^6(x)dx\\ \\ \int [si{n}^2(x){]}^{3}dx\\ \\ \int [\frac{1-cos(2x)}{2}{]}^{3}dx\\ \\ \frac{1}{8}\int 1-3cos(2x)+3co{s}^2(2x)-co{s}^3(2x)dx\end{array}$$

So, we can reduce the powers. 1 can integrate directly. 3cos^2 is another even power one. cos^3 is like the odd power one above.

combination of sin & cosine

$$\begin{array}{r}\int si{n}^3(x)co{s}^2(x)dx\\ \\ \int si{n}^2(x)co{s}^2(x)sin(x)dx\\ \\ \text{b/c we peeled off sin, that will be the u-substitution}\\ \\ \int (1-co{s}^2(x))co{s}^2(x)sin(x)dx\\ \\ u=cos(x)\\ \\ du=-sin(x)dx\\ \\ -\int (1-u^2)u^2\cdot -du\\ \\ -\int u^2-u^{4}du\\ \\ -\frac{1}{3}{u}^3+\frac{1}{5}{u}^5+C\\ \\ -\frac{1}{3}co{s}^3(x)+\frac{1}{5}co{s}^5(x)+C\end{array}$$