Integral

[Loki123]

So I put this example in an integral calculator and it gave me besides the arcsin answer I got, the arctan answer. I got curious so I tried to get it, but I couldn't get it exactly the same. I am missing the minus and the fraction is flipped. Any advice?

 

[BigBeachBanana]

Do you know the standard integral of arcsin(x)?

 

[Deleted member 4993]

So I put this example in an integral calculator and it gave me besides the arcsin answer I got, the arctan answer. I got curious so I tried to get it, but I couldn't get it exactly the same. I am missing the minus and the fraction is flipped. Any advice? View attachment 32725

Did you "differentiate" your answer and "differentiate" calculator-answer to check whether those are different and where is the difference?

 

[Steven G]

Why do you have to get that answer?

Also, \(\displaystyle \dfrac{\cos a}{\sqrt{\cos^2 a}}\neq 1\)

Why would you think it is 1?

 

[Dr.Peterson]

So I put this example in an integral calculator and it gave me besides the arcsin answer I got, the arctan answer. I got curious so I tried to get it, but I couldn't get it exactly the same. I am missing the minus and the fraction is flipped. Any advice? View attachment 32725

Are you asking how to get the second form by a different approach to the integral; or how to transform your (much nicer) form into it?

I see no reason to want to do the former, apart from, as you say, curiosity.

The first and third forms you show are indeed equivalent, as you can see by graphing. The form you say you "need" to get differs by a constant. which is closely related to the reciprocal identity ...

 

[Loki123]

Do you know the standard integral of arcsin(x)?

I don't think so

 

[Loki123]

Why do you have to get that answer?

Also, \(\displaystyle \dfrac{\cos a}{\sqrt{\cos^2 a}}\neq 1\)

Why would you think it is 1?

Cosx/Cosx=1

 

[Dr.Peterson]

Also, \(\displaystyle \dfrac{\cos a}{\sqrt{\cos^2 a}}\neq 1\)

Why would you think it is 1?

Cosx/Cosx=1

It's true only when cos(x) is positive; you failed to say anything about the range of your variable alpha. In fact, if you defined it correctly, you can be sure the cosine is positive, so this is not the error in your work. But it is important to define substitutions fully.

Have you been able to prove my claim about the three expressions? It's a valuable skill.

 

[Steven G]

Cosx/Cosx=1

Yes, but that is not what you had. sqrt(cos^2t) is not cos t. Try replacing cos t with -5, ie does (sqrt(-5))^2 = -5???

 

[Loki123]

It's true only when cos(x) is positive; you failed to say anything about the range of your variable alpha. In fact, if you defined it correctly, you can be sure the cosine is positive, so this is not the error in your work. But it is important to define substitutions fully.

Have you been able to prove my claim about the three expressions? It's a valuable skill.

well i know it's positive because
x^2=t=sqrt(3)cosa

 

[BigBeachBanana]

I don't think so

Consider the standard integral of [imath]\arcsin(u)[/imath], which is
[math]\arcsin(u)=\int \frac{1}{\sqrt{1-u^2}}\,du[/math]Notice your problem resembles it (in red).
[math]\int\frac{x}{\sqrt{3-x^4}}\,dx=\int\frac{x}{\sqrt{3}\sqrt{1-\frac{x^4}{3}}}\,dx=\int\frac{x}{\sqrt{3}}\cdot \red{\frac{1}{\sqrt{1-\frac{x^4}{3}}}}\,dx[/math]If you recognized that, it's an obvious choice to make a u-subsitution [imath]u^2=\frac{x^4}{3}[/imath] or [imath]u=\frac{x^2}{\sqrt{3}}[/imath].
I'll let you work out the details but clearly, this isn't something you're expected to do and I second Dr.P's comments. It's more valuable for you.

 

[Dr.Peterson]

well i know it's positive because
x^2=t=sqrt(3)cosa

Really? You said

​

This implies that the sine is positive, but not necessarily the cosine!

What you really need to do (and I won't claim I always do it carefully) is to define what your alpha is. You defined t as x^2, so we know it has only one value (and is positive). But you have not said how to obtain alpha, just that it is some angle with a certain sine.

Probably your intent is to think of alpha as the inverse sine of t/sqrt(3), which implies it is between -pi/2 and pi/2. That implies that the cosine is positive. So you're right, but for the wrong reasons, and without having provided proper grounds for the conclusion.