Substitution Definition and 797 Threads

This is part of the working from f(3x^2-1)^2xdx; I don't understand from when 6x becomes 1/6

I have ##1-x^2 = 1- \sin^2 θ = \cos^2 θ## and ## dx =cos θ dθ## ##\int_0^{0.5} (1-x^2)^{1.5} dx = \int_0^{\frac{π}{6}} [cos ^2θ]^\frac{3}{2} dθ = \int_0^{\frac{π}{6}} [cos ^4θ] dθ## Suggestions on next step.

First, we rewrite the term ##|\vec r-\vec r_q|## in the following way: $$|\vec r-\vec r_q|= \sqrt{(\vec r-\vec r_q)^2} = \sqrt{\vec r^2 + \vec r_q^2 -2\vec r\cdot\vec r_q} = \sqrt{r^2 + r_q^2 -2rr_q\cos\theta}$$ Due to rotational symmetry, we go to spherical coordinates: $$\phi_{e;\vec r_q} =...

Here's the problem: ##\int_0^{2\pi} \cos^{-1}(\sin(x)) \mathrm{d}x## If I do the substitution ##u = \sin(x)##, both the limits of integration become 0 and the integral would result in 0. But the graph of the function tells me the area isn't 0. Where am I going wrong?

I know how to solve similar ODEs like ## \frac{\partial^2 x}{ \partial t^2} + b \frac{\partial x}{ \partial t} + C x =0 ## Where one can let ## x = e^{rt}##, and the equation becomes ## r^2 + b r + C =0 ## Which can be solved as a quadratic equation. But now the problem is that there is...

How did you find PF?: Looking for Circuits on Pirani Gauges I am in the process of building a High Vacuum system and obtained an outdated Pirani Gauge controller TM120 on ebay. The unit is from Leybold Heraeus company and is built with quality. I contacted them on information on the manuals...

Method 1, Pretty straightforward, $$\int_{-1}^0 |4t+2| dt$$ Let ##u=4t+2## ##du=4 dt## on substitution, $$\frac{1}{4}\int_{-2}^2 |u| du=\frac{1}{4}\int_{-2}^0 (-u) du+\frac{1}{4}\int_{0}^2 u du=\frac{1}{4}[2+2]=1$$ Now on method 2, $$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} |4t+2|...

Hi, With respect to the techniques mentioned in point 2 and 3: Can someone explain or even better, post a link for an explanation or a videos showing the use of these two techniques. Below excerpt shows problems 4 and 5 referenced in the above 2 points:

We can solve ##y'(x) = (ax+b)y(x)## by rearranging to obtain ##\frac{y'}{y} = ax +b## and solving in terms of an exponential. I tried an alternative technique to see if it would work, and somewhere I went wrong. The point of the technique is that a slightly simpler version of the problem should...

Last night I tried to calculate from an automatically generated Wolfram Alpha problem set: $$\int{\frac{1}{\sqrt{x^2+4}}}dx$$ I answered $$\ln({\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}})+C$$ The answer sheet gave: $$\ln({\sqrt{x^2+4}+x})+C$$ I couldn't see where I had gone wrong, so I tried...

Given is a function ##P(E)## and its derivative ##f(E)##. Writing ##E## in terms of ##v## according to ##E=\frac{1}{2}mv^2## gives the derivative ##g(v)=f(E)mv## and ##dE=mvdv##. My issue arises from the premise that I learned; Integrals and derivatives are based on steps of a fixed constant...

Evaluate ##\displaystyle\int_{0}^{3}\frac{x+3}{\sqrt{x^{3}+1}}dx+5## W|A returned 11.7101 ok subst is probably just one way to solve this so ##u=x^{3}+1 \quad du= 3x^2##

1.\[ \int x^2 e^{x^3} dx \] 2. \[ \int sin(2x-3)dx \] 3. \[ \int (\cfrac {3dx}{(x+2)\sqrt {x^2+4x+3}} ) \] 4. \[ \int (\cfrac {x^3}{(x^2 +4)^\cfrac {3}{2}} )dx \]

Is it possible to solve this integral? I think the substitution ##x=-u## does not help at all since it only changes variable ##x## to ##u## without changing the integrand much. Using that substitution: $$\int \frac{6x^2+5}{1+2^x}dx=-\int \frac{6u^2+5}{1+2^{-u}}du$$ Then how to continue? Thanks

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Can someone please tell me the book that contain integration using hyperbolic substitution for beginner? I know that hyperbolic functions is taught in Calculus book but most of them is only some identities and inverses of hyperbolic functions.

Can someone please show me an example of integration using hyperbolic substitution? Thank you

Question: Helmholtz as defined in text: My attempt so far

My considers a type of differential equation $$\frac{\mathrm{d} y}{\mathrm{d} x} = f\left(\frac{y}{x} \right )$$ and proposes that it can be solved by letting ##v(x) = \frac{y}{x}## which is equivalent to ##y = xv(x)##. Then it says $$\frac{\mathrm{d} y}{\mathrm{d} x} = v + x\frac{\mathrm{d}...

a) I think I got this one right. Please let me know otherwise We have (let's leave the ##x## dependence of the fields implicit :wink:) $$\mathscr{L} = N \Big(\partial_{\alpha} \phi \partial^{\alpha} \phi^{\dagger} - \mu^2 \phi \phi^{\dagger} \Big) = \partial_{\alpha} \phi^{\dagger}...

I'm really stuck on this one, I was able to get the answer but not by the substitution method. So its the weight as A and B so I get A + B = 24 A(3) = B(5) so in my head I calculate a few pairs, 3 x 5 = 15 but 3 + 5 only = 8 so the next pair would be 10 and 6 which is still to small so I move...

Hi, I've a doubt about the applicability of the substitution theorem in circuit theory. Consider the following picture (sorry for the Italian inside it :frown: ) As far I can understand the substitution theorem can be applied to a given one-port element attached to a port (a port consists of...

I completely forgot how to solve these so here's my attempt: $$z = au + bv$$ $$z = a(x^2 + y^2) + be ^{-x^2/2}$$ $$z'_x = 2ax - bxe ^{-x^2/2}$$ $$z'_y = 2ay$$ Put that into the original equation and you get $$y * (2ax - bxe ^{-x^2/2}) -x * (2ay) = $$ $$-ybe^{-x^2/2} = xyz$$ $$z =...

Evaluate $\displaystyle\int{\dfrac{{(1-\ln{t})}^2}{t} dt=}$ $a\quad {-\dfrac{1}{3}{(1-\ln{t})}^3+C} \\$ $b\quad {\ln{t}-2\ln{t^2} +\ln{t^3} +C} \\$ $c\quad {-2(1-\ln{t})+C} \\$ $d\quad {\ln{t}-\ln{t^2}+\dfrac{(\ln{t^3})}{3}+C} \\$ $e\quad {-\dfrac{(1-\ln{t^3})}{3}+C}$ ok we can either expand...

Problem: Consider the equation $$\frac{\partial v}{\partial t} = \frac{\partial^{2} v}{\partial x^2} + \frac{2v}{t+1}$$ where ##v(x,t)## is defined on ##0 \leq x \leq \pi## and is subject to the boundary conditions ##v(0,t) = 0##, ##v(\pi, t) = f(t)##, ##v(x,0) = h(x)## for some functions...

Hello, In high school, I had been taught about finding substitution resistance from Wheatstone bridge. The formula: a. If the cross product of ##R1## and ##R3## is same as ##R2## and ##R4##, the galvanometer in the middle (##R_5##) can be omitted and use series-parallel principle to solve for...

Would this be valid manipulation for ##x\in[0,\,\pi/2]##? I know that it is integrable by parts, I just want to know where did the manipulation become invalid, if it did, and why. Thank you! $$\begin{align*} \mathrm I&=\int_a^b x^2\sin2x\,dx\\ &\text{I know that...

Not sure how do I start from here, but do I let $$u = lnx$$ and substitute? Cheers

ok this is from my overleaf doc so too many custorm macros to just paste in code but I think its ok,,, not sure about all details. appreciate comments... I got ? somewhat on b and x and u being used in the right places

AP Calculas Exam Problem$\textsf{Using $\displaystyle u=\sqrt{x}, \quad \int_1^4\dfrac{e^{\sqrt{x}}}{\sqrt{x}}\, dx$ is equal to which of the following}$ $$ (A)2\int_1^{16} e^u \, du\quad (B)2\int_1^{4} e^u \, du\quad (C) 2\int_1^{2} e^u \, du\quad (D) \dfrac{1}{2}\int_1^{2} e^u \, du\quad...

##132,289≡1973* 67 + 98## ##1973≡98*20+13## ##98≡13*7+7## ##13≡7*1+6## ##7≡6*1+1## now in reverse my attempt is as follows, ##1≡7-6## ## 1≡7-(13-7)## ##1≡2*7-(1973-20*132,289+1340*1973)## ##1≡2*7-(1341*1973-20*132,289## which is correct but my interest is in finding the inverse of 1973 help?

Hello everyone. First off, I'm sorry if this post is excessively long, but after tackling this for so many hours I've decided I could use some help, and I need to show everything I did to express exactly what I wish to do. Also, to be clear, this post deals with integration by substitution. Now...

When substitution is properly used for a set of equations, I believe you get a new equation with solutions that are also solutions of both of the previous equations. The following equation has solutions x = 0 and x = 1. ##x=x^2## This next equation has solutions x = -2 and x = 2. ##x^2=4##...

Hello, For my homework I am supposed to get- into the form of a Bessel equation using variable substitution. I am just not sure what substitution to use. Thanks in advance.

This is the integral I try to take. ##\int\sqrt{1+9y^2}## and ##9y^2=tan^2\theta## so the integral becomes ##\int\sqrt{1+tan^2\theta}=\sqrt {sec^2\theta}##. Now I willl calculate dy. ## tan\theta=3y ## and ##y=\frac {tan\theta}3## and ##dy=\frac{1+tan^2\theta}3## Here is where I can only...

Homework Statement Find an equation that defines IMPLICITLY the parameterized family of solutions y(x) of the differential equation: 5xy dy/dx = x2 + y2 Homework Equations y=ux dy/dx = u+xdu/dx C as a constant of integration The Attempt at a Solution I saw a similar D.E. solved using the y=ux...

Consider a reaction: H2+CuCl2= Cu+2HCl This is a substitution reaction.But is this may not be a reversible reaction since Cu is less active than .So Cu can't substitute H from HCl and make a backward reaction.Is my thinking right?

Homework Statement Homework Equations So the question is asking to solve an integral and to use the answer of that integral to find an additional integral. With part a, I don't have much problem, but then I don't know how to apply the answer from it to part b. I know I should subsitute all...

Homework Statement Integrate: $$\int \frac{dx}{x^2\sqrt{4-x^2}}dx$$ Homework EquationsThe Attempt at a Solution I got to the final solution ##\int \frac{dx}{x^2\sqrt{4-x^2}}dx=-\frac{1}{4}cot(arcsin(\frac{1}{2}x))##. But It's the method where you transform that to the solution...

Hi! I am trying to solve problems from previous exams to prepare for my own. In this problem I am supposed to find the improper integral by substituting one of the "elements", but I don't understand how to get from one given step to the next. Homework Statement Solve the integral by...

$\textsf{Evaluate the integral}$ $$I=\displaystyle\int\frac{x^2}{\sqrt{9-x^2}}$$ $\textit{from the common Integrals Table we have}$ $$\displaystyle I=\int\frac{u^2}{\sqrt{u^2-a^2}} \, du =\frac{u}{2}\sqrt{u^2-a^2}+\frac{a^2}{2} \ln\left|u+\sqrt{u^2-a^2}\right|+C$$...

What I have done: I changed all fractions to common denom and that gave me 5y-5x=1 (1) *I numbered the fractions 5y+2x=5 (2) Then: 5y=5-2x Substitute into equation 1 (5-2x)-5x=1 5-7x=1 x=4/7 Thing is my answer says I should be getting x=0 Any hints?

Hi members, See attacged PDF file for my question Thank you

I have seen two approaches to the method of integration by substitution (in two different books). On searching the internet i came to know that Approach I is known as the method of integration by direct substitution whereas Approach II is known as the method of integration by indirect...

Homework Statement ## \int {sin} \frac{\pi x} {L} dx ##Homework Equations u substitution The Attempt at a Solution If i make ## u = \frac{\pi x} {L} ## and then derive u I get ## \frac {\pi}{L} ## yet the final solution has ## \frac {L}{\pi} ## The final solution is ## \frac {L}{\pi} - cos...

Hi, I posted a question here a few days ago regarding some questions I've been doing on an online quiz. I seem to be getting stuck on the integral substitution questions. I've been slowly making progress, but some of these questions have been confusing me, and reading up on them is only giving...

Homework Statement $$y'=-\frac{1}{10}y+(cos t)y^2$$ when doing substitute for ##z=\frac{1}{y}## I understand this is ##z(t)=\frac{1}{y(t)}## I know t is independent variable and y is dependent variable but I want to know what is z role here, is it change the dependent variable? when...

Hi, I've got this problem that I've been trying to work out. I think most of my problems come from the fact that I am not yet well versed in u substitution when it comes to integrals. I'm also not 100% sure what the problem is asking. I've tried doing a couple of things, but they don't seem to...

Homework Statement :[/B] I was learning the use of standard forms in method of substitution in solving integration. My book has given this method for solving integrals of the type ##\int \frac{lx +m}{ax^2+bx+c} dx##: As an example, the book gives this one: Homework Equations :[/B] The...

This is a small part of a question from the book, so I think the format does not really apply here. When doing questions for solving differential equation with substitution, I encountered a substitution ## y(x)=\frac{1}{v(x)} ##. And I am not sure about the calculus in finding ## \frac{dy}{dx}...