# Help request for calculating an integral

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.

• 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 $x$
– Masd
Apr 12 at 16:11
• you can't do that $${a+b\over ac+b}\neq{1\over c}$$, besides you lost the integral sign
– Masd
Apr 12 at 16:14
• It might be easier to just substitute as step 1.
– Mike
Apr 12 at 16:27

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}

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 $$2\sin t\cos t$$?

• Thanks for your help. What do you mean by identities? Apr 12 at 16:39
• $(\sin t + \cos t) ^2 = 1 + \sin 2t$ Apr 12 at 16:41
• I mean trigonometric identities. Masd's answer shows what I was getting at. Apr 12 at 16:47

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