Integration into Inverse trigonometric functions using Substitution
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- Опубликовано: 21 авг 2024
- This calculus video tutorial focuses on integration of inverse trigonometric functions using formulas and equations. Examples include techniques such as integrating by substitution, u-substitution, splitting fractions and completing the square. This video contains plenty of practice problems of integrating inverse trig functions.
Arc Length Problems:
• Arc Length Calculus Pr...
Surface Area Problems:
• Surface Area of Revolu...
Work Problems - Calculus:
• Work Problems - Calculus
Integration By Parts:
• Integration By Parts
Tabular Method:
• Integration By Parts -...
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Trigonometric Integrals:
• Trigonometric Integrals
Trigonometric Substitution:
• Trigonometric Substitu...
Integration By Partial Fractions:
• Integration By Partial...
Trapezoidal Rule:
• Trapezoidal Rule
Simpson's Rule:
• Simpson's Rule & Numer...
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Improper Integrals:
• Improper Integrals - C...
Integration Into Inverse Trig:
• Integration into Inver...
Integration of Rational Functions:
• Integration of Rationa...
Integral of Logarithmic Functions:
• Integral of Logarithmi...
Integrating Exponential Functions:
• Integrating Exponentia...
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May you be willing to do a video about hyperbolic functions and their inverse?
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Hey so at 13:50 why if you have (x/x^2+1) did you have the whole denominator U instead of applying the arctan formula? X^2 is U^2 and 1 is a^2.
Also as everyone else says you are the goat at math.
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I think it's because when u^2 = x^2, u = x, du = dx. Therefore you will not to be able to let the numerator (x) disappear, and get the integral in the right form. If u^2 would have been x^4, it would have been another story since then you can multiply by du/2x. (like an example in the vid)
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Professor Organic Chemistry Tutor, thank you for an outstanding video/lecture on Integration into Inverse Trigonometric Functions using Substitution. Integration into Inverse Trigonometric Functions ranges from simple to complex, however pattern recognition also helps when computing these complex Integrals in Calculus. This is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
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31:30 can also use natural log formula du/u since the 2 terms in the denominator are perfect squares, so it will become xdx / 5 - (x² - 4) or -x² + 9.. we can let u = -x² + 9 and du = -2x dx, and dx = du/ -2x... We now have integral of x du/-2x in the numerator, we can cancel x then put the constant -1/2 outside so we now have -1/2 integral of du/u which is just -1/2 ln | -x² + 9 | + c
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These questions can also be solved using trigonometric substitution, I'm getting a little confused please can someone tell me when we'll use trigonometric substitution and when this inverse thing
Mohammad Hamza bro I was thinking the same thing
@@thomascruickshank7407 Bro I was thinking the same thing. Currently in Calc 2
correct me if i am wrong, but i think you can use this method if the functions are in the denominator, while in trig substitution, the functions are in the numerator or there is simply no denominator.
To state the truth you can actually do it by any method because integration doesn't have discrete methods like differentiation 😊
@mohammed aswr. Its confusing. Dunno when to use either of them
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Tip:
When you derive x^n and the equation is
×dx/something you could simply over the derivative of it with only just the number so it will be xdx=du/n. No need to bring the x so there will be no cancellation.
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will you please make a video about hyperbolic functions like sinh and cosh and those formulas ;-; thank you for everything
There are also inverse hyperbolic substitution formulas; arccosh, arcsinh, arcsech, arctanh
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there are somethings that u forgot to apply just like 1/a at 31:01 in this video, but it's alright u r a legend
the answer must be 3/2 arctan x+3/ 2 + c (the 2nd term of the answer)
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in 19: 42 you have a solution to the problem. but will the problem also be correct if you use the same u substitution for both integrals, thus getting the answer 1/2 ln (u) - 3ln(u) + c? or must you always apply the the rules to each integral separately? thus you must apply arc tan rule at the second part of the integral?
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im sure this is obvious, but why are we only using the principle square root? Is it because of the domain of the function? If so, then can't one of the terms, like the x term, be negative?
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Who has seen this. In 13.25. That problem of integral of x÷(×square +1)dx . He used u substitution. And that problem looks exactly like the question given which was integral of x÷(x rest to 4+36). So y dint he use u substitution at the beginning of the question given....
The problem at the beginning had an x^3 on the numerator. So we could not use U substitution, which is the reason he performed long division in the first place. Let me know if you are still confused?
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