Post by: DeviantPain12 on March 22, 2019, 06:13:18 pm
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?
Post by: AlphaZero on March 22, 2019, 09:14:49 pm
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 \(\displaystyle \int \arcsin(x)\,\text{d}x\).
Post by: DeviantPain12 on March 22, 2019, 10:52:19 pm
Post by: schoolstudent115 on March 22, 2019, 10:56:44 pm
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: ]=\frac{x}{\sqrt{1-x^2}}+arcsin(x))
Spoiler
Integrating both sides yields: =\int{\frac{x}{\sqrt{1-x^2}}}+\int{arcsin(x)})
So, dx}=xarcsin(x)-\int{\frac{x}{\sqrt{1-x^2}} dx})
To solve the integral on the RHS, let
Substituting in:
Therefore, dx}=xarcsinx-\sqrt{1-x^2}+C)
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) +t(x))
, where )
is an arbitrary function obtained through differentiation, you can define:
=\int{t(x) dx})
, so )
.
You now have (using equation (1)):=\int{t(x) dx}=\int{\frac{df}{dx}-\lambda(x) dx})
, which can usually be solved with normal integration.
If you are told to integrate some 'target' function
You now have (using equation (1)):
Post by: AlphaZero on March 23, 2019, 01:30:36 pm
...
Just take a bit more care with negative signs. The correct answer is \[\int\arcsin(x)\,\text{d}x=x\arcsin(x)+\sqrt{1-x^2}+C\]