does anybody have an advice how I can integrate this $$\int\frac{e^{\arcsin x} + x + 1}{\sqrt{1-x^2}}dx$$ I tried substitution $\arcsin x=t$, but was not able to finish it.
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What are you missing, this change of variable makes it trivial ? – Feb 24 '15 at 11:22
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The substitution should help... – Panglossian Oporopolist Feb 24 '15 at 11:22
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"I tried substitution arcsinx=t, but was not able to finish it" We really must know where you were stopped, to answer meaningfully your question. – Did Feb 24 '15 at 11:27
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Yeah, I already got it. Sorry, for trivial question. – dimaastronom Feb 24 '15 at 11:33
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@YvesDaoust it really does) – dimaastronom Feb 24 '15 at 11:36
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@Kugelblitz it did) – dimaastronom Feb 24 '15 at 11:37
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@Did passed it already, anyway thx – dimaastronom Feb 24 '15 at 11:37
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Alright. No problem :) – Panglossian Oporopolist Feb 24 '15 at 11:37
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Remember, a problem is trivial to some, difficult to others, so don't out yourself down. It's an interesting question for math amateurs like me nonetheless! – Panglossian Oporopolist Feb 24 '15 at 11:38
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@Kugelblitz thank you for the support. I agree with you ;P – dimaastronom Feb 24 '15 at 11:42
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I got this. When you substitute $arcsinx=t$ you got really good output. $$\int\frac{e^{arcsinx}}{\sqrt{1-x^2}}dx$$ if $arcsinx=t$ => $dt=\frac{dx}{\sqrt{1-x^2}}$ and the integral transforms into $$\int e^tdt=e^{arcsinx} + C_1$$ The whole answer is $$e^{arcsinx} -\sqrt{1-x^2} + arcsinx +C$$ Thats it.
dimaastronom
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