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I have a problem with this integral

$$\int_\ \frac{\cos x }{\sin x \sqrt{1+\cos^2x}} \, dx$$

Using substitution $u = \sin x $ we get

$$\int_\ \frac{1 }{\ u \sqrt{2-u^2}} \, du$$

I think the next step is to use simple fractions, but I don't know how to do it.

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  • $\begingroup$ Try letting $u=\sqrt t$ $\endgroup$ – user170231 Jan 23 '16 at 8:20
  • $\begingroup$ I don't understand your idea. Could you explain this ? $\endgroup$ – davoid Jan 23 '16 at 9:20
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Hint

Multiply both numerator and denominator by $u$ and then use the substitution $2-u^2=t^2.$ To get $$\int \frac{u}{u^2\sqrt{2-u^2}} \, du=\int \frac{1}{t^2-2} \, dt.$$ Now you can use partial fractions.

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  • $\begingroup$ You don't even really need partial fractions, you'll just get an answer in terms of an inverse hyperbolic tangent function. $\endgroup$ – Tom Jan 3 '17 at 10:39

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