Help with solving integrals?

Question

anonymous · Answer

$$\int\limits_{1}^{-1} \frac{ e^{\tan^{-1}(y) } }{ 1+y^2 } dy$$

anonymous · Answer

the 1 and -1 are the wrong way on the integral whoops!

tkhunny · Answer

$u = \tan^{-1}(y)$  You can't let ones like this get away.

anonymous · Answer

I got that part.. but I don't know where to go after $$\int\limits_{-1}^{1} \frac{ e^u }{ du }$$

tkhunny · Answer

How did the "du" get in the denominator?

anonymous · Answer

let 
$$\Large  x=\tan^{-1}(y)$$
$$\Large  dx=\frac{1}{1+y^2}dy$$
$$\Large \int\limits_{}^{}  e^{x}dx$$

anonymous · Answer

okay! so is that the final answer do do I need to evaluate it at 1 and -1?

anonymous · Answer

I'm sorry. No. no no no. You need to change your limits of integration.

tkhunny · Answer

You have a choice:

1) Change the limits with your substitution, or
2) Work the indefinite integral and substitute back to the variable where you started before applying the limits.

anonymous · Answer

integral limits will also change 

when 
y=1
x=tan^-1(1)=pi/4
when y=-1
x=tan^-1(1)=-pi/4

integral becomes
$$\Large \int\limits_{-\pi/4}^{\pi/4}e^xdx$$

anonymous · Answer

$$e^u|_{-\pi/4}^{\pi/4} 

= e ^{\pi/4} - e ^{-\pi/4}$$  ?

tkhunny · Answer

Or, like I said, switch back to the original variable, before the substitution, and don't change the limits.

$e^{\tan^{-1}(y)}|_{-1}^{1}$

It is the same.  With a little practice, it will often be clear which is easier.