# Integral question

1. Apr 17, 2007

### 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

plz help

#### Attached Files:

• ###### tan4.GIF
File size:
102.1 KB
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2. Apr 17, 2007

### 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.

3. Apr 18, 2007

### 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

???

4. Apr 18, 2007

### StatusX

Because cos^2+sin^2=1.

5. Feb 11, 2009

### transgalactic

but the "u" expression represents x

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

??

6. Feb 11, 2009

### HallsofIvy

Staff Emeritus
No, u= sin(arctan(x))

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.

7. Feb 11, 2009

### 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?