[tex] 2u du=e^{x} dx [tex]
I finally get to
so if this is right
can somebody help me with hte second part of the last integral
and if it's wrong somekind of solution
i get something different for the integral. by the way, i don't like the substitution you made because , but anyway, you'd get something like if we say
an alternative is to use two substitutions. first and then and you'll get a similar integral to the one above
if you're still interested, completing the square coupled with substitution can take care of your original integral
Hello, akhayoon!
Your substitution should have come out like Jhevon's . . .
Let
Substitute: .
Back-substitute: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Here's another method . . . it's either brilliant or incredibly stupid.
Let
Substitute: .
Back-substitute
We have: .
This is a right triangle with: .
Hence: .
So we have: .
. .
. .
You mission (should you decide to accept it)
. . is to prove that my two answers are equivalent.
.
i say brilliant! so crazy, it's brilliant! ...though not for the faint of heart, or those new to the world of integrals. how did you discover such a substitution? was the integral given in a "special form"?
laterYou mission (should you decide to accept it)
. . is to prove that my two answers are equivalent.
.
One can play with trig. sub. as many times as necessary.
700% personal opinion, I don't like trig. substitutions, some times it turns the integrand into a mess, but it depends of the problem.
I prefer workin' with algebra, you can get faster solutions, especially the reciprocal substitution, which can turn a nasty integral, into a cute one. (Not always, of course.)