Helping you with the integral of 1/(x^2*(x^2+1)^(3/2)) using trigonometric substitution, Calculus 2

Learn how to use trigonometric substitution for the integral of 1/(x^2*(x^2+1)^(3/2)). This question is from Threads: www.threads.ne...
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[@kingbeauregard]

About the "+ C", maybe teach from early days that the solution to every antiderivative has two parts: the homogeneous solution and the particular solution. For antiderivatives, the homogeneous solution is always "C", since it answers the question: "what, if we take the derivative of it, is always zero?" But we're typically more interested in the particular solution, so we spend all our time working on just the particular solution. At the very end, the complete answer is the sum of the particular solution and the homogeneous solution, and THAT is why we include the "+ C" only at the end.

[@MikehMike01]

Telling students to only put +C at the end implies it’s wrong to put them earlier… if you want to be technical, the integrals along the way could be +C1 +C2 etc. and the final answer +Cn… I think that should help people understand better

[@kingbeauregard]

@@MikehMike01 I'm thinking more, let's be clear about what we're doing: we're focusing on finding the particular solution, and only adding in the homogeneous solution at the very end. Then it's not just a matter of completeness vs sloppiness, it's a deliberate, mathematically justified approach.
It will also make things feel real familiar when you get to differential equations.

[@blueslime5855]

​@@MikehMike01they are just the same constant though since you aren't adding anything to them no? You simply transformed the equation. I believe you only change the constant if you have some kind of constant like 1 leftover which is redundant to keep

[@MikehMike01]

@@blueslime5855 they are arbitrary integers which sum to another arbitrary integer

[@ThePowerfulOne07]

My calculus professor once said that Calculus is 90% algebra/geometry/trig and you only learn 10% of actual calculus. Still amazes me to this day!

[@kingbeauregard]

The way my high school calculus teacher put it, most calculus problems are just one line of calculus and the rest is algebra. Of course that was beginning calculus, but the concept holds.

[@ThePowerfulOne07]

@@kingbeauregard yes it does

[@AverageCommentor]

if you clean up the result at the end you'll get -(2x^2+1)/(x*sqrt(x^2+1)) + C

[@bjornfeuerbacher5514]

Or "simply" use the substitution u = x/sqrt(x²+1), then you will arrive directly at the integral at 4:30. ;)

[@kingbeauregard]

Not really a strategy you can teach though.

[@alex_ramjiawan]

Explain please.

[@bjornfeuerbacher5514]

@@alex_ramjiawan Explain how I got that idea? Or explain how to do the calculation?

[@alex_ramjiawan]

​@@bjornfeuerbacher5514 The calculation with your substitution.

[@bjornfeuerbacher5514]

@@alex_ramjiawan First, solving u = x/sqrt(x²+1) for x², you get x² = u²/(1-u²). Second, using the quotient rule, you get du/dx = 1/sqrt(x²+1)³. Putting both results into the original integral transforms it into the integral (1-u²)/u² du.

[@d-hat-vr2002]

In order to have the integrand after the x = tan θ substitution represent the same function as the original integrand, we can notice that the domain of the original integrand is all x ∉ 0. Constraining the allowable values of θ to -π/2 < θ < π/2 ∉ 0 gives the same range of y values.
In other words, the integrand written at 1:41 has the same area under curve as the original integrand, but with the domain -π/2 < θ < π/2 ∉ 0.
For the multiplication of powers at 2:02 to be correct, the base (sec θ in this case) must be positive, which it is in this question since sec θ is positive over the whole interval -π/2 < θ < π/2 ∉ 0. On the other hand, this would not be correct if the substitution x = tan θ were made under the assumption that the new domain would be 0 < θ < π .

[@eboaly]

ur videos carried my math grade thank you so much🙏🙏🙏

[@Mediterranean81]

1/x*sqrt(x^2+1) + c

[@bachvaroff]

Forget'bout trig's, substitute $1 + x^2 = t^2 x^2 \\ x^2 = \frac{1}{t^2 - 1} \\ x^{-2} + 1 = t^2 \\ -2 x^{-3} dx = 2t dt$ etc. etc. 🙂

[@r1ckthe]

can you help me doing the integral of v(arctanv) dv ?? pls i need help

[@cyrusyeung8096]

Try integration by parts (differentiate arctan(v) and integrate v)

[@bjornfeuerbacher5514]

Is this supposed to mean v times the arctan of v? Strange use of parentheses. :/

[@r1ckthe]

@@bjornfeuerbacher5514 yep, varctanv dv

[@bprpcalculusbasics]

I got you, ua-cam.com/video/iaDrbWG0zDo/v-deo.html

[@leonardobarrera2816]

How I can ask?
Plz BPRP, answer me

[@MikehMike01]

No one uses threads lmao dead app

[@Mephisto707]

Forget threads man, nobody is there, go X and support free speech.

[@nakoiz4851]

could you use u=secθ for 2:30