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I am trying to calculate the following inntegral $$\int \frac{2\sqrt{1- x^2}}{2 x\sqrt{1- x^2}+ 5}\, dx. $$ My attempt:
$$\int \frac{2\sqrt{1- x^2}}{2 x\sqrt{1- x^2}+ 5}\, dx = \frac{1}{x} - \int \frac{5}{2 x^2 \sqrt{1- x^2}+ 5x}\, dx, $$ and putting $x = \sin t$ the integral becomes $$\int \frac{5 \cos t \, dt}{2\sin^2 t \cos t + 5\sin t} $$ Here I have faced with the problem!

Thanks for any helps.

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  • $\begingroup$ I believe you've done it incorrectly on you first step you divided $$2\sqrt{1-x^2}+5\over 2x\sqrt{1-x^2}+5$$ look at the $x$ $\endgroup$
    – Masd
    Apr 12 at 16:11
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    $\begingroup$ you can't do that $${a+b\over ac+b}\neq{1\over c}$$, besides you lost the integral sign $\endgroup$
    – Masd
    Apr 12 at 16:14
  • $\begingroup$ It might be easier to just substitute as step 1. $\endgroup$
    – Mike
    Apr 12 at 16:27

3 Answers 3

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Substitute $x=\frac t{\sqrt{1+t^2}}$

\begin{align} \int \frac{2\sqrt{1- x^2}}{2 x\sqrt{1- x^2}+ 5}dx =& \int \frac{2}{(1+t^2)(5t^2+2t +5)}dt\\ =&\ \int \frac{t+\frac25}{t^2+\frac25 t +1}-\frac t{1+t^2}\ dt \end{align}

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Your initial simplification is wrong. Implementing the trig substitution right away, while assuming $\cos y>0$ for brevity, yields

$$\int \frac{2\sqrt{1- x^2}}{2 x\sqrt{1- x^2}+ 5} \, dx \stackrel{x=\sin t}= \int \frac{2\cos^2t}{2\sin t \cos t+5} \, dt$$

Hint for proceeding: what identities do you know that can be applied to $\cos^2t$ and $2\sin t\cos t$?

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  • $\begingroup$ Thanks for your help. What do you mean by identities? $\endgroup$
    – Mary_26
    Apr 12 at 16:39
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    $\begingroup$ $(\sin t + \cos t) ^2 = 1 + \sin 2t$ $\endgroup$
    – Mary_26
    Apr 12 at 16:41
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    $\begingroup$ I mean trigonometric identities. Masd's answer shows what I was getting at. $\endgroup$
    – user170231
    Apr 12 at 16:47
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Letting $x=\sin u $ we get: $$ \int \frac{2\cos^2 u}{2\sin u \cos u+5} du $$ Note that $2\cos^2 u =\cos(2u)+1$ and $2\sin u \cos u =\sin(2u)$ $$ \int {\cos(2u)+1\over\sin(2u)+5} du = \frac{1}{2}\int \frac{\cos(v)+1}{\sin(v)+5}dv \tag{$v=2u$} $$ This leaves you with a nice and clean integral, this is your hint

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