How to Write arcsin in Log Form for Integration

[azatkgz]

It seems to that i went wrong way.

Homework Statement

\int \frac{arcsin(e^x)dx}{e^x}

The Attempt at a Solution

=-\frac{arcsin(e^x)}{e^x}+\int \frac{dx}{\sqrt{1-e^{2x}}}


\int \frac{dx}{\sqrt{1-e^{2x}}}=\frac{x}{\sqrt{1-e^{2x}}}+\int \frac{xe^{2x}dx}{(1-e^{2x})^{\frac{3}{2}}}

 

[ppyadof]

Ever thought about writing arcsin in log form (the inverse function of sin when sin is written in complex numbers), that way something might happen between the logs and the exp's. I haven't tried it, so I don't know, its just a suggestion.

 

[azatkgz]

Here is the another way I tried.
u=e^{-x}
-\int arcsin(\frac{1}{u})du=-\int arccsc(u)du=-uarccsc(u)-\int\frac{du}{\sqrt{u^2-1}}

 

[azatkgz]

Than for
u=sec\theta\rightarrow du=\frac{sin\theta d\theta}{cos^2\theta}

 

[azatkgz]

\int\frac{du}{\sqrt{u^2-a}}=-\int\frac{d\theta}{cos\theta}
But I don't know,actually,how to integrate last one.

 

[rocomath]

-\int\frac{d\theta}{\cos{\theta}}

trig identity

-\int\sec{\theta}d\theta

 

[azatkgz]

Ok,than please check everything.
\int \frac{arcsin(e^x)dx}{e^x}

u=e^{-x}

-\int arcsin(\frac{1}{u})du=-\int arccsc(u)du=-uarccsc(u)-\int\frac{du}{\sqrt{u^2-1}}




u=sec\theta\rightarrow du=\frac{sin\theta d\theta}{cos^2\theta}

=-uarccsc(u)+\int\sec{\theta}d\theta

=-uarccsc(u)+ln|sec\theta+tan\theta|

=-e^{-x}arccsc(e^{-x})+ln|e^{-x}+e^{-x}\sqrt{1-\frac{1}{e^{-2x}}}|

 

[azatkgz]

Sorry,I've posted answer to another question here.

 

[Gib Z]

Post 7, you made a sign error. Suddenly the sec integral becomes positive...other than that, the answer is fine, but it would look nicer if you took the e^(-x) under the sqrt sign in the log, as the argument of the log is always positive anyway, no need for absolute value signs.