What is the integral,
$$\int\frac{dx}{x + \sqrt{1-x²}}\ ?$$
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$\begingroup$ Is my edit correct? If not please roll back. $\endgroup$– DavidCommented May 27, 2015 at 5:53
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$\begingroup$ Hmm, wonder if it helps to rationalize the denominator by multiplying and dividing by $x-\sqrt{1-x^2}$? $\endgroup$– MPWCommented May 27, 2015 at 6:03
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$\begingroup$ There is a classical change of variable suggested by the radical. $\endgroup$– Claude LeiboviciCommented May 27, 2015 at 6:09
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1 Answer
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$$ \int\frac{dx}{x + \sqrt{1-x^2}}\ $$ Take $x=\sin\theta$, $$ \int\frac{\cos\theta}{\sin\theta + \cos\theta} \, d\theta $$ $$ \frac{1}{2}\int\frac{2\cos\theta}{\sin\theta + \cos\theta} \, d\theta $$ $$ \frac{1}{2}\int \frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta} \, d\theta+\frac{1}{2} \int \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\, d\theta $$ $$ \frac{1}{2}\theta + \frac{1}{2}\ln|\cos\theta+\sin\theta|+C $$ Substitute $\theta=\sin^{-1}x$ and you will get the answer.
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$\begingroup$ It'd be better if constant C is added. $\endgroup$ Commented May 27, 2015 at 6:13