I need help on two questions on integration of inverse trig functions?

Question

I need the integrals of
a) (e^x sec e^x tan e^x) dx / (( 36-(sec e^x)^2)^.5)

and

b) e^(2ln arctan x) dx / (1+x^4)

Thanks to you both. I will tell my professor. ^^

Omg I forgot that d(sec e^x) =
sec e^x*tan e^x*e^x. The e^x was what I was worrying about. ^^

Answer 1

a)
substitute y = sec(e^x) => dy = e^x·sec(e^x)·tan(e^x) dx
∫ e^x·sec(e^x)·tan(e^x) / √(36 - sec(e^x)²)) dx
= ∫ 1/√(36 - y²) dy
next substitute y = 6·sin(z) => dy = 6·cos(z) dz
= ∫ 6·cos(z) dz/√(36 - 6·sin²(z)) dz
= ∫ cos(z) dz/√(1 - sin²(z)) dz
= ∫ cos(z) dz/√(cos²(z)) dz
= ∫ 1 dz
= z + c
= arcsin(y/6) + c
= arcsin(sec(e^x)/6) + c

(another solution, which is found by substitution
y = 6·cos(z) , is:
∫...dx = -arccos(sec(e^x)/6) + c )


b)
∫ e^{ 2·ln(arctan(x))} / (1+x^4) dx
= ∫ e^{ ln([arctan(x)]²)} / (1+x^4) dx
= ∫ [arctan(x)]² / (1+x^4) dx

So far so good, but i don't know how to solve this integral.

Maybe there a is typo. If the denominator would be (1+x²)
you would found a solution by substituting
z = arctan(x) => dx = (1+x²) dz :
∫ [arctan(x)]² / (1+x²) dx
= ∫ z² dx
= (1/3)·z³ + c
= (1/3)·[arctan(x)]³ + c

But for the given nominator (1+x^4) i don't have a clue to solve.

Answer 2

Schmiso did a fine job on the first problem.
The second one is not elementary. In fact,
I ran it through the Wolfram integrator and it
couldn't find a formula for it.