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Find the $\space \displaystyle\int e^{\sin^{-1}x}~\mathrm dx$ .

I started by making a substitution. Let $u=\sin^{-1}x$, and so one can conclued that:

$\begin{align}1)&\mathrm du=\displaystyle\frac{1}{\sqrt{1-x^2}}\mathrm dx\\2)&x=\sin u \end{align}$

So, the integral stays:

$\begin{align}\int e^{\sin^{-1}x}\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \mathrm dx &=\int e^u\sqrt{1-\sin^2u}~\mathrm du=\int e^u\sqrt{\cos^2u}~\mathrm du=\int e^u\cos u~\mathrm du\end{align}$

Now, I tryed integration by parts but I could't managed. In Wolfram there is a complicated formula that I never heard about. Is there an intuitive way to finish this integral? Thanks.

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4 Answers 4

up vote 2 down vote accepted

Two consecutive integrations by parts should give you an equation satisfied by the primitive. More specifically, $$ \int_{-\infty}^a e^u\cos u\,\mathrm du = e^a\cos a + \int_{-\infty}^a e^u\sin u\,\mathrm du.$$

A second integration by parts gives an equation for $\int_{-\infty}^a e^u\cos u\,\mathrm du$.

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I set $\mathrm dv=\cos u$ and $s=e^{u}$. And so, $v=\sin u$ and $\mathrm ds=e^{u}$.Then $e^{u}\sin x-\int e^{u}\sin x$ –  João Jun 18 '13 at 20:44
    
Same thing, really; it works just as well. I just did my integration by parts with $\mathrm dv=e^u$ and $s=\cos u$. –  jathd Jun 18 '13 at 20:48
    
But if I continuing integrate by parts, it will be an integral left to integrate by parts. It seems it would never end. –  João Jun 18 '13 at 20:53
    
And will the remaining integral be? –  jathd Jun 18 '13 at 21:01
    
After two integrations by parts the remaining integral will be equal to the original. –  João Jun 18 '13 at 21:02

You can integrate the last integral twice by parts, then solve for it.

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One can call the following "Cheating" or "Method of Undetermined Coefficients."

Guess that the answer will look like $Ae^u\cos u+Be^u\sin u$.

Differentiate. We get $$Ae^u(\cos u -\sin u) +Be^u(\sin u+\cos u).$$ In order for the above to be $e^u\cos u$ we need $A+B=1$, and $-A+B=0$. Solve. We get $A=B=\frac{1}{2}$, so our integral is $$\frac{1}{2}e^u\cos u+\frac{1}{2}e^u\sin u+C.$$

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+1 for "Cheating" or "Method of Undetermined Coefficients" –  ronno Jun 18 '13 at 21:40

Let $I$=$\displaystyle \int e^{\displaystyle x}\cos x \space \mathrm dx\ $. Integrating $I$ by parts gives $$\begin{align} u=\cos x \; & \implies \frac{\mathrm du}{\mathrm dx}=- \sin x \\ \frac{\mathrm dv}{\mathrm dx}=-\sin x \; & \implies v=e^{\displaystyle x} \\ \end{align}$$ Hence $\displaystyle\space I=uv-\int v \cdot \mathrm du=e^{\displaystyle x}\cos x +\int e^{\displaystyle x}\sin x \space\mathrm dx\ $. Now let $\displaystyle I_1=\int e^{\displaystyle x}\sin x \space\mathrm dx\ $. $$\begin{align} u_1=\sin x \; & \implies \frac{\mathrm du_1}{\mathrm dx}= \cos x \\ \frac{\mathrm dv_1}{\mathrm dx}=e^{\displaystyle x} \; & \implies v_1=e^{\displaystyle x} \\ \end{align} $$ Hence $\displaystyle \space I_1=u_1v_1-\int v_1 \cdot \mathrm du_1=e^{\displaystyle x}\sin x +\int e^{\displaystyle x}\cos x \space \mathrm dx$

Now $$\begin{align} I &= e^{\displaystyle x}\cos x+(e^{\displaystyle x}\sin x +\int e^{\displaystyle x}\cos x \space \mathrm dx)\\ & = e^{\displaystyle x}(\cos x + \sin x) - I\\ \text{Hence } \space 2I =e^{\displaystyle x}(\cos x + \sin x) \\ &\\ \text{Finally, } \space \;I &=\frac{e^{\displaystyle x} (\cos x + \sin x)}{2}=\int e^{\displaystyle x}\cos x \space \mathrm dx\ \end{align} $$

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