Question

This is the question of integration by parts and it is killing me

Integration of e^(m* inverse of sinx) dx

Answer 1

\[\int\limits e ^{m \sin^{-1} x} dx\]

Answer 2

use \[\sin^{-1}x=t \]

Answer 3

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.

Answer 4

=> x=sint
=> dx=cost dt
so the expressio becomes
\[\int\limits e^{mt}costdt\]
now use integration by parts

Answer 5

I tried everything but the solution which I am getting is not matching the solution given in the textbook

Answer 6

@ vamgadu I tried but the answer is no matching the texttbook answer

Answer 7

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

Answer 8

\[
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}
\]