Struggling with an Integral? Here's How to Solve It!

[transgalactic]

i am stuck on the last part of it?

i added a file the question and the way i tried to solve it

 

[HallsofIvy]

Your problem, then, is sin(arctan(x))? One way to do that is to use trig identities to write sine as a function of tangent. I think it is much easier to go back to their initial definitions. Draw a right triangle having angle \theta, "opposite side
" of length x, and "near side" of length 1 so that tan(\theta)= x/1= x and, of course, \theta = arctan(x). Now use the Pythagorean theorm to find the length of the hypotenuse. What is sin(x)= "opposite side/hypotenuse"

 

[transgalactic]

i tried to solved it by the triangle method.

can you please show this triangle
i tried to emagine this.


you said one side is 1 the other is x
so the hypotenuse is (1+x^2) ^0.5

its not halping me in finding

sin(arctan(x))

?

 

[HallsofIvy]

Yes, if one side is 1, the other x, the the hypotenuse is \sqrt{1+ x^2}.
My point was that since you have set it up so that tan(\theta)= x, then \theta= arctan(x). That means sin(arctan(x))= sin(\theta). Now, what is sin(\theta) in that triangle?

 

[transgalactic]

i got that
sin(teta)=x/(1+x^2)^0.5


is it correct??


how i solve it in the other method?

 

[Gib Z]

The other way is ugly. Let x= tan u. Then dx = sec^2 u du. The integral then becomes
\int \frac{\sin u}{\cos^2 u} du where you then let y= cos u.

 

[Gib Z]

The other way is longer and, if I was marking it and I saw this solution next to the one Halls pointed you to, I would say this solution is more of a mechanical labor task rather than mathematical insight when compared to the other one. But it's always good to know how to do it in more ways than one, so here it is:

First let u= tan x. Then in the resulting integral, let y= cos u. (or if you want to think of it this way, one big substitution of y= cos (tan x) )

 

[transgalactic]

can i solve all the variation of this type of question
like
sin(arcctg(x))
in the triangle method??

 

[Gib Z]

Yes! Its wonderful! In that case, the triangle method gives \frac{1}{\sqrt{1+x^2}}.

 

[transgalactic]

cool thank you very much