Integration question (1 Viewer)

[Slidey]

We sub x = atan u in, and in the end we sub tan u = (x/a) and sec = sqrt (x^2+a^2) /a in to get a function in term of x. We dont use arc function
the substitution x = asec u doesnt simplify the integral so i dont think that we can use it. Maybe its possible but i havent tried yet

You misunderstood. You'll need to use inverse sec to get back to a form in x at the end of the question. (I'm guessing).

Otherwise pretend I'm talking about a question which requires you to use inverse sec to get back to a function in x.

 

[ngogiathuan]

You misunderstood. You'll need to use inverse sec to get back to a form in x at the end of the question. (I'm guessing).

Otherwise pretend I'm talking about a question which requires you to use inverse sec to get back to a function in x.

We let x = a tan u right, therefore x/a = tan u, and sec u = sqrt ( x^2 + a^2). Thats all. Once u get the function in sec u and tan u just sub these values in to get a function in x.
Yeah im still taking my time to comprehend ur post just because there are too many exponential and complex number so it looks kinda scareeeee
Well thanks a lot for ur replies

 

[Slidey]

We let x = a tan u right, therefore x/a = tan u, and sec u = sqrt ( x^2 + a^2). Thats all. Once u get the function in sec u and tan u just sub these values in to get a function in x.

Oh, yeah I'm fully aware of how to solve this integral.

Suppose you use the substitition x=tan(u) for some integral. Then you get an answer to an integral of something in the form ln(sec(u)) or somesuch. In order to get this back to a function in x, you need to use the inverse sec function. That's basically all I am saying. And because you use the inverse sec function, have you introduced some restrictions on what values x can take?

 

[ngogiathuan]

i understand what you're saying but in this case we dont need to use inverse sec function. For example, i have ln(sec u + tan u) right, i want to get back to x, so just sub in tan u = x/a and sec u = sqrt (x^2 + a^2/a) it becomes ln((x+sqrt(x^2+a^2))/a)

 

[Slidey]

Oh! Very good. Sorry, I totally missed that.

I was noting how you also get that answer if you use inverse sec:
tan(u)=x/a
opposite side length x
adjacent length a
hypotenuse length = sqrt(x^2+a^2) (take positive root because length is positive)
cos(u)=a/sqrt(x^2+a^2)
sec(u)=sqrt(x^2+a^2)/a

Was your question why do we say the answer is ln((x+sqrt(x^2+a^2))/a) instead of ln((x-sqrt(x^2+a^2))/a)? Using inverse sec you can answer that by saying side lengths are always positive. Using your way however... not sure.

 

[ngogiathuan]

Oh! Very good. Sorry, I totally missed that.

I was noting how you also get that answer if you use inverse sec:
tan(u)=x/a
opposite side length x
adjacent length a
hypotenuse length = sqrt(x^2+a^2) (take positive root because length is positive)
cos(u)=a/sqrt(x^2+a^2)
sec(u)=sqrt(x^2+a^2)/a

Was your question why do we say the answer is ln((x+sqrt(x^2+a^2))/a) instead of ln((x-sqrt(x^2+a^2))/a)? Using inverse sec you can answer that by saying side lengths are always positive. Using your way however... not sure.

Yep thats exactly right!!!
The only thing about this working is by letting u = an angle in a right-angled triangle, you already restricted u from 0 to pi/2, which is not the range satisfied the condition that x go from -infinity to +infinity
Like Affinity said, if u restrict u from -pi/2 to pi/2, sec u will definitely > 0, so u can take +ve square root without messing anything up