Login

Welcome, Guest. Please login or register.

March 29, 2024, 01:59:27 am

Author Topic: Integral of Arcsin Necessary?  (Read 1308 times)  Share 

0 Members and 1 Guest are viewing this topic.

DeviantPain12

  • Fresh Poster
  • *
  • Posts: 4
  • Respect: 0
Integral of Arcsin Necessary?
« on: March 22, 2019, 06:13:18 pm »
0
Hey all
In my Cambridge textbook it says "The antiderivative of sin−1 is not required for this course, but the area can still be determined as follows."

Does this mean it's not part of the study design and won't show up in the exam? I ask this because I've seen it say specifically (for another topic) that it's not part of the study design. Do they mean the same thing?

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 352
  • \[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\]
  • Respect: +160
Re: Integral of Arcsin Necessary?
« Reply #1 on: March 22, 2019, 09:14:49 pm »
+1
In an exam, you will not be asked to integrate any of the inverse circular functions without any help.

However, they could ask you to use integration by recognition. Here's an example question.

Question 1 (3 marks)

a.     Find \(\dfrac{d}{dx}\!\Big[x\arcsin(x)\Big]\).

b.     Hence, find \(\displaystyle \int \arcsin(x)\,\text{d}x\).
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Mathematical Sciences Diploma, University of Melbourne


DeviantPain12

  • Fresh Poster
  • *
  • Posts: 4
  • Respect: 0
Re: Integral of Arcsin Necessary?
« Reply #2 on: March 22, 2019, 10:52:19 pm »
0
Ah gotcha. Thanks!

schoolstudent115

  • Trendsetter
  • **
  • Posts: 117
  • Respect: +6
Re: Integral of Arcsin Necessary?
« Reply #3 on: March 22, 2019, 10:56:44 pm »
0
In an exam, you will not be asked to integrate any of the inverse circular functions without any help.

However, they could ask you to use integration by recognition. Here's an example question.

Question 1 (3 marks)

a.     Find \(\dfrac{d}{dx}\!\Big[x\arcsin(x)\Big]\).

b.     Hence, find \(\displaystyle \int \arcsin(x)\,\text{d}x\).

Just so I know, would this be how you'd be expected to find that integral:
a.
Spoiler
Use product rule to find that the derivative is:
b.
Spoiler
Integrating both sides yields:
So,
To solve the integral on the RHS, let
Substituting in:

Therefore, 

*Note for similar integration questions*:
Spoiler
A generalised method would be as such:
If you are told to integrate some 'target' function , and you know that some (1) , where is an arbitrary function obtained through differentiation, you can define:
, so .

You now have (using equation (1)): , which can usually be solved with normal integration.
« Last Edit: March 22, 2019, 11:14:38 pm by schoolstudent115 »
2021: ATAR: 99.95
2022-2024: University of Melbourne, BSci (Major in Mathematics and Statistics)

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 352
  • \[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\]
  • Respect: +160
Re: Integral of Arcsin Necessary?
« Reply #4 on: March 23, 2019, 01:30:36 pm »
0
...

Just take a bit more care with negative signs. The correct answer is \[\int\arcsin(x)\,\text{d}x=x\arcsin(x)+\sqrt{1-x^2}+C\]
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Mathematical Sciences Diploma, University of Melbourne