u-substitution

I believe this is a mechanism of applying the [Chain rule](20220326132748-derivatives.org::*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?

&\frac{1}{2}\int \frac{x^2}{\sqrt{x^2-1}} 2x dx\\ &\text{Because we know that $u=x^2+1$, $x^2=u-1$..}\\ &\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{aligned} $$ ### 4 use the trig identity: $\tan^2(x) = \sec^2(x)-1$ $$ \begin{aligned} &\int \tan^2(x) dx \\ &\int \sec^2(x)-1 dx \\ &\int \sec^2(x) dx - \int 1 dx \\ &tan(x) - x \end{aligned} $$ ## Differenation? - [x] DONE Completed: 2022-04-23 In https://www.youtube.com/watch?v=OKY6m9x3fME, you mention that $[sin(x^2)]' = \cos(x^2) \cdot 2x$. Where did that 2x come from? ### Q: Is integration the opposite of differentation? differentiation = find derivative integration = anything with the $\int$ sign. aka antiderivatives ### Q: I seem to be missing up when to do derivatives vs antiderivatives. Any pointers? derivative is the slope of the tangent line of the graph integral is the area underneath the curve. ## u-substitution w/ definite integrals https://www.youtube.com/watch?v=r4Gjl7GNM7I When dealing w/ definite integrals, when you've gone from $dx$ to $du$, you have to change the range of the definite integral. One mechanism of doing that is to substitute in the a and b values into the u function for x to get the new value. ### 1 $$ \begin{aligned} &\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{aligned} $$ ### 2 $$ \begin{aligned} \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 du dx\\ \int_{\frac{\sqrt{3}}{2}}^1 &u^2 \cdot du dx\\ &= \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{aligned} $$ ## Integrating powers of trig functions https://www.youtube.com/watch?v=tlKBHHGfIso We can solve powers of trig functions w/ substitutions like $f(x) = \sin^m(x)\cos^n(x)$. - $\sin^2(x) + \cos^2(x) = 1$ - $\sin^2(x) = \frac{1-\cos(2x)}{2}$ - $\cos^2(x) = \frac{1+\cos(2x)}{2}$ Useful b/c they're reducing the power of the sin/cos. $$ \begin{aligned} \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{aligned} $$ ### Odd powers & pythagoriean identity Odd powers are easier w/ Pythagoriean identity. $$ \begin{aligned} &\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}sin^3(x)+\frac{1}{5}\sin^5(x)+C \end{aligned} $$ ### Even powers and half-angle identities e.g. $\int sin^6(x)dx = \int[sin^2(x)]^3 dx$ Even powers usually require a bunch more work. $$ \begin{aligned} \int sin^6(x) dx\\ \int [sin^2(x)]^3 dx\\ \int [\frac{1-cos(2x)}{2}]^3 dx\\ \frac{1}{8} \int 1-3cos(2x)+3cos^2(2x)-cos^3(2x) dx \end{aligned} $$ So, we can reduce the powers. 1 can integrate directly. 3cos^2 is another even power one. cos^3 is like the odd power one above. ### combination of sin & cosine $$ \begin{aligned} \int sin^3(x) cos^2(x) dx\\ \int sin^2(x) cos^2(x) sin(x) dx\\ \text{b/c we peeled off sin, that will be the u-substitution}\\ \int (1-cos^2(x)) cos^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}cos^3(x) ~~ \frac{1}{5}cos^5(x) ~~ C \end{aligned} $$ ### I don't see any homework problems around trig identity problems. Where can I find some to practice? - [x] DONE Completed: 2022-05-11 ahem. found them.