Evaluation of $\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$

Question

Evaluation of $$\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$

$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$

Put $x=\cos 2 \theta\;,$ Then $dx = -2\sin 2 \theta d\theta$ and Changing Limit, We get

$$I = \int_{0}^{\frac{\pi}{4}}\frac{2\sin 2 \theta}{\sqrt{2}\cos \theta+\sqrt{2}\sin \theta+2}d\theta$$

So $$I = \int_{0}^{\frac{\pi}{4}}\frac{\sin 2 \theta}{\cos\left(\theta-\frac{\pi}{4}\right)+1}d\theta$$

Now Put $\displaystyle \theta-\frac{\pi}{4}=t\;,$ Then $d\theta = dt$ and changing limits

So $$I = \int_{-\frac{\pi}{4}}^{0}\frac{\cos 2t}{\cos t+1}dt=2\int_{-\frac{\pi}{4}}^{0}\frac{2\cos^2 t-2+1}{\cos t+1}dt$$

after that we can solve easily,

My question is can we solve it without Using Trig substution,

Plz explain me

Thanks

This could be tricky; for example $1+(1+\sqrt{1-x})(1+\sqrt{1+x})=2+\sqrt{1-x}+\sqrt{1+x}+\sqrt{1-x^{2}}$ — , May 13 '16 at 10:48
The integral is solved by Wolfram Alpha, but it's not trivial. By the way, $$I=2\sqrt{2}-1-\frac{\pi}{2}$$ — Yuriy S, May 13 '16 at 10:51

score 6 · Accepted Answer · answered May 13 '16 at 12:19

I'm not sure this is the most straight forward way, but here it is anyway.

Start with the substitution $\sqrt{1-x}\mapsto x $. The integral transforms into

$$I=\int_0^1 \frac{2 x}{2+x+\sqrt{2-x^2}}dx.$$ Next, rationalize the denominator by multiplying the numerator and the denominator by $2+x-\sqrt{2-x^2}.$
This turns out to work very well, because $\,\,(2+x)^2-(2-x^2)=2(1+x)^2.$ So we get that the integral equals $$I=\int_0^1 \frac{x(2+x-\sqrt{2-x^2})}{(1+x)^2}dx.$$ Finally, integrate by parts (note that $\displaystyle \frac1{(1+x)^2}=-\frac{d}{dx} \frac1{1+x}$): $$I=-\frac{x(2+x-\sqrt{2-x^2})}{1+x}\Bigg{|}_0^1+\int_0^1 2\left(1+\frac{x-1}{\sqrt{2-x^2}}\right)dx \\\\=-1+2-2\left(\sqrt{2-x^2}+\sin^{-1}\frac{x}{\sqrt{2}}\right)\Bigg{|}_0^1 \\\\=2\sqrt{2}-1-\frac{\pi}{2}.$$

I think this is the best way. +1 – Yuriy S May 13 '16 at 12:27 — Yuriy S, May 13 '16 at 12:27
Bravo! - An excellent choice of substitution at the outset. – May 13 '16 at 14:28 — , May 13 '16 at 14:28

score 2 · Answer 2 · answered May 13 '16 at 12:18

multiply the integrand by

$$ 1=\frac{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)}{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)} $$

doing the alegbra correctly this indeed eliminates the roots in the denominator and we end up with $$ I=\frac{1}{2}\int_{0}^{1}\frac{-\sqrt{1-x} x+\sqrt{1+x} x-2 \sqrt{1-x}+2 \sqrt{1-x^2}-2 \sqrt{x+1}+2}{x^2}dx=\\ $$

taking into account a proper limiting procedure for $x\rightarrow 0,1$, this integral can be split in elementary pieces (solvable by integration by parts and/or subs of the form $y\rightarrow y\pm 1$)

Evaluation of $\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$

2 Answers2