This is the question of integration by parts and it is killing me Integration of e^(m* inverse of sinx) dx
\[\int\limits e ^{m \sin^{-1} x} dx\]
use \[\sin^{-1}x=t \]
Integration of parts is so fun though =/ remember the acronym LIPET for what to use as your "u" in your integration by parts Logs Inverses Polynomials Exponential Trig In this case, separate it first into e^m * e^inverse sinx only thing that trips me up here is that you integrating with respect to x but where does m come from....I'd just assume it to be a constant and therefore you can do the backwards power rule to antiderive it.
=> x=sint => dx=cost dt so the expressio becomes \[\int\limits e^{mt}costdt\] now use integration by parts
I tried everything but the solution which I am getting is not matching the solution given in the textbook
@ vamgadu I tried but the answer is no matching the texttbook answer
i agree with vamgadu ^ do you get this answer http://www.wolframalpha.com/input/?i=integrate+e^%28mt%29+cos%28t%29+dt then you have to plug sin(x) in for t
\[ dv = dx \\ u= e^{m \sin ^{-1}(x)}\\ v= x\\ du=\frac{m e^{m \sin ^{-1}(x)}}{\sqrt{1-x^2}}\\ u v - \int v du =x e^{m \sin ^{-1}(x)}-\frac{m \left(m x-\sqrt{1-x^2}\right) e^{m \sin ^{-1}(x)}}{m^2+1}=\\ \frac{\left(m \sqrt{1-x^2}+x\right) e^{m \sin ^{-1}(x)}}{m^2+1} \]
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