Simplifying a Complicated Integral: Tips and Tricks

[transgalactic]

i added a file with of the integral
and how i "solved" it

on the final stage of my integral i gut stuck on a
complicated fuction

is there a way to change it so it will look simpler

please help

 

[StatusX]

So you're asking if you can simplify \sin(2\tan^{-1}(x))? Use the double angle formula to get it in terms of \sin(\tan^{-1}(x)) and \cos(\tan^{-1}(x)). Then if you put u=\sin(\tan^{-1}(x)), you have \cos(\tan^{-1}(x))=\sqrt{1-u^2} and:

\frac{u}{\sqrt{1-u^2}}=\frac{\sin(\tan^{-1}(x)}{\cos(\tan^{-1}(x))}=\tan(\tan^{-1}(x))=x,

which you can solve for in terms of u.

 

[transgalactic]

you said u=sin(arctan(x))
how come i have cos(arctan(x))=(1-u^2)^0.5

i didnt understand what to do step by step

?

 

[StatusX]

Because cos^2+sin^2=1.

 

[transgalactic]

but the "u" expression represents x

i heard of some triangle method of solving stuff like
<br /> \sin(2\tan^{-1}(x))<br />

??

 

[HallsofIvy]

but the "u" expression represents x

No, u= sin(arctan(x))

i heard of some triangle method of solving stuff like
<br /> \sin(2\tan^{-1}(x))<br />

??

For something simple, like sin(tan^-1(x)), you can imagine a triangle with "opposite side" x and "near side" 1 so that "opposite side over near side" = tan(angle)= x and angle= tan^-1(x). Then sin(tan^-1(x))= sin(angle) which is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenus: \sqrt{x^2+ 1} and you get sin(tan^-1(x))= x/\sqrt{x^2+ 1}.

However, the "2" multiplying tan^-1(x) makes that much harder.

 

[transgalactic]

how to build the new integral
i need to build a du for that which i the derivative of
u= sin(arctan(x))

what is the full new integral?