Integrating Exponentials with Roots that have Roots? (And other small Q's)

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kape
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Hello, I have a few questions! I need clarification on certain points that were not very clear in my calculus book.

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Question 1:

I know that [tex]\int e^{ax} dx = \frac{1}{a} e^{ax}[/tex]

But how do you integrate [tex]\int e^{ax^2} dx[/tex]?


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Question 2:

I know that integrating by parts is [tex]\int (something) dx= uv - \int vdu[/tex]

But what if there is a range?

If it is [tex]\int_{a}^{b} (something) dx[/tex] does it equal [tex]\left[ uv \right]_{a}^{b} - \int_{a}^{b} vdu[/tex] or does it simply equal [tex]uv - \int_{a}^{b} vdu[/tex]?


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Question 3:

How do you integrate [tex]\int log_ax dx[/tex] and [tex]\int e^{ln|secx|} dx[/tex].

In fact, is [tex]e^{ln|secx|}[/tex] reducable?


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Question 4:

I was taught that arcsinx exist only in the range [tex]\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right][/tex] and [tex]\left[ \frac{\pi}{2}, \frac{3\pi}{2} \right][/tex] (I think because it fails the horizontal test if it isn't in those ranges)

If so, is it possible to integrate [tex]\int_{0}^{\pi} xarcsinx dx[/tex]? (If it is possible, is it because it isn't simply arcsinx but xarcsinx?)


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Question 5:

I am having a lot of problems integrating fuctions with exponents etc that have complex roots. My elementary calculus is shaky at best and I'm taking Advanced Engineering Mathematics (Kreyzig) - I have to. Can anyone recommend me any links or books that may help me?


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Reply to HallsofIvy:

Thank you for your reply! I have a question about your reply on question 1: In my Adv Eng Maths (Kreyzig) book, one of the questions is how to integrate [tex]\int xe^{x^2/2}[/tex] and the answer is [tex]e^{x^2/2} + C[/tex] but I don't understand how to do it!
 
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Answer to question 1: You don't. That integral,
[tex]\int e^{ax^2}dx[/tex]
is well known not to have an elementary integral. In fact, precisely that integral (with a= -1) is important in Statistics and it's integral is defined to be "Erf(x)", the error function.

If it were
[tex]/int xe^{ax^2}dx[/tex]
then you could make the substitution [itex]u= e^{ax^2}[/itex] and have the xdx already for du= 2xdx.

Question 2, Yes, just plug the limits of integration into the formula.

Question 3 seems to have disappeared.
 
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Notice that [tex]\frac{d}{dx}\left(\frac{x^2}{2}\right)=x[/tex]
So it is actually [tex]\int{f'\left(x\right)e^{f\left(x\right)}=e^{f\left(x\right)}+c[/tex]
 
Thank you for answering questions 1 & 2.. I think I understand.

Sorry question 3 was deleted, don't quite know how that happened.

Also, I have one more question: (which is kind of similar to question 3)


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Question 6

How do you integrate:

[tex]\int \frac{1}{x^a} dx[/tex]

[tex]\int \frac{1}{a^x} dx[/tex]

[tex]\int \frac{1}{a^{bx}} dx[/tex]

[tex]\int \frac{1}{a^{bx^{c}}} dx[/tex]

Should I have learned this somewhere? I don't see these in the integral tables or rules..
 
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[tex]\frac{1}{x^a}= x^{-a}[/tex]
Use the power rule.

[tex]\frac{1}{a^x}= a^{-x}[/tex]
Make the substitution u= -x and then use
[tex]\int a^x dx= \frac{a^x}{ln a}[/tex]

Same thing:
[tex]\frac{1}{a^{bx}}= a^{-bx}[/tex]
Make the substitution u= -bx.

[tex]\int \frac{1}{a^{bx^c}}dx[/tex]
depends strongly on what c is. There is no general anti-derivative.