0
$\begingroup$

Evaluate $$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$

My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what indicated that said substitution would work etc.)

My initial substitution was $x= \sin \theta$ which tidies it up a bit

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \space dx$$

Also the answer from Wolfram Alpha is;

$$ = \frac{1}{4}\bigg( log(1-2 x^2)+2 \tanh^{-1}\bigg(\frac{x}{\sqrt{1-x^2}}\bigg)+2 \sin^{-1}x\bigg)+constant$$

I need to be able to answer questions like this in under 15 mins so any help is appreciated

$\endgroup$
0
$\begingroup$

BIG HINT:

$$\int\frac{1}{x+\sqrt{1-x^2}}\space\text{d}x=$$


Substitute $x=\sin(u)$ and $\text{d}x=\cos(u)\space\text{d}u$.

Then $\sqrt{1-x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\arcsin(x)$:


$$\int\frac{\cos(u)}{\sin(u)+\cos(u)}\space\text{d}u=$$ $$\int\frac{\sec^3(u)}{\sec^3(u)}\cdot\frac{\cos(u)}{\sin(u)+\cos(u)}\space\text{d}u=$$ $$\int\frac{\sec^2(u)}{\sec^2(u)+\sec^2(u)\tan(u)}\space\text{d}u=$$


Prepare to substitute $s=\tan(u)$. Rewrite $\frac{\sec^2(u)}{\sec^2(u)+\sec^2(u)\tan(u)}$ using $\sec^2(u)=1+\tan^2(u)$:


$$\int\frac{\sec^2(u)}{1+\tan(u)+\tan^2(u)+\tan^3(u)}\space\text{d}u=$$


Substitute $s=\tan(u)$ and $\text{d}s=\sec^2(u)\space\text{d}u$:


$$\int\frac{1}{s^3+s^2+s+1}\space\text{d}s=$$ $$\int\left(\frac{1-s}{2(s^2+1)}+\frac{1}{2(s+1)}\right)\space\text{d}s=$$ $$\int\frac{1-s}{2(s^2+1)}\space\text{d}s+\int\frac{1}{2(s+1)}\space\text{d}s=$$ $$\frac{1}{2}\int\frac{1-s}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$ $$\frac{1}{2}\int\left(\frac{1}{s^2+1}-\frac{s}{s^2+1}\right)\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$ $$-\frac{1}{2}\int\frac{s}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$


Substitute $p=s^2+1$ and $\text{d}p=2s\space\text{d}s$:


$$-\frac{1}{4}\int\frac{1}{p}\space\text{d}p+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$ $$-\frac{\ln\left|p\right|}{4}+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$ $$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$


Substitute $w=s+1$ and $\text{d}w=\space\text{d}s$:


$$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{1}{2}\int\frac{1}{w}\space\text{d}w=$$ $$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{\ln\left|w\right|}{2}+\text{C}$$

$\endgroup$
  • $\begingroup$ you substituted sec^2 for sec^3 in your second substitution $\endgroup$ – KingJ Dec 5 '15 at 21:53
  • $\begingroup$ No I substituted $s=\tan(u)$ $\endgroup$ – Jan Dec 5 '15 at 21:54
  • $\begingroup$ you had sec^3 du and went to ds but sec^2 du = ds? $\endgroup$ – KingJ Dec 5 '15 at 21:56
  • $\begingroup$ @RSparkes maybe now it is clear for you! $\endgroup$ – Jan Dec 5 '15 at 22:02
  • $\begingroup$ Thank you for fixing the typo $\endgroup$ – KingJ Dec 5 '15 at 22:03
7
$\begingroup$

There's a simple solution somewhere in M.SE which goes like this...

$$I = \int_0^{\pi/2} dt \frac{\cos{t}}{\cos{t}+\sin{t}} $$ $$J = \int_0^{\pi/2} dt \frac{\sin{t}}{\cos{t}+\sin{t}} $$

$$I + J = \frac{\pi}{2}$$ $$I - J = \int_0^{\pi/2} dt \frac{\cos{t}-\sin{t}}{\cos{t}+\sin{t}} = \int_0^{\pi/2} dt \frac{(\cos{t}+\sin{t})'}{\cos{t}+\sin{t}}=[\log{(\cos{t}+\sin{t})}]_0^{\pi/2} = 0$$

Thus $I = J = \frac{\pi}{4}$.

ADDENDUM

The above technique requires the easy computation of the antiderivative, which you should see is

$$\frac12 \left [t + \log{(\cos{t}+\sin{t})} \right ] + C = \frac12 \left [\arcsin{x} + \log{(x+\sqrt{1-x^2})} \right ] + C$$

$\endgroup$
3
$\begingroup$

Another simple solution:

\begin{align} &x\mapsto\sin u\\ &I=\int_0^{\pi/2} \frac{\cos u}{\sin u + \cos u}du\\ &u\mapsto \frac{\pi}2-v\\ &I=\int_0^{\pi/2} \frac{\sin v}{\sin v + \cos v}dv\\ &\therefore 2I=\int_0^{\pi/2} \frac{\sin u + \cos u}{\sin u + \cos u}du=\frac{\pi}{2}\\ &\therefore I=\frac{\pi}{4} \end{align}

$\endgroup$
1
$\begingroup$

Since no one used it, let me consider the case of the antiderivative $$I=\int\frac{\cos x}{\sin x + \cos x} \space dx$$ Now, use the tangent half-angle substitution (Weierstrass substitution) $t=\tan(\frac x 2)$.

We so obtain $$I=\int \frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}\space dt$$ But $$t^4-2 t^3-2 t-1=(t^2+1)(t-r_1)(t-r_2)\qquad r_{1,2}=1\pm \sqrt 2$$ Using partial fraction decomposition, we have $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1-t}{t^2+1}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ that is to say $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1}{1+t^2}-\frac{2t}{1+t^2}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ and each term is easy to integrate. After simplifications, this leads to $$I=\tan ^{-1}(t)+\frac{1}{2} \log \left(\frac{-t^2+2 t+1}{t^2+1}\right)$$ Now, if integration is from $t=0$ to $t=1$, the logarithms disappear and youare just let with $\tan ^{-1}(1)=\frac \pi 4$.

$\endgroup$
-1
$\begingroup$

Finding indefinite integral:

\begin{align} &x\mapsto\sin u\\ &I=\int \frac{\cos u}{\sin u + \cos u}du\\ &=\frac12\int \frac{\cos u - \sin u + \cos u + \sin u}{\sin u + \cos u}du=\frac12 \int\frac{\cos u-\sin u}{\sin u + \cos u}du+\frac u2\\ &=\frac12 \ln (\cos u + \sin u) + \frac u2+C\\ &=\frac12 \ln (\sqrt{1-x^2} + x) + \frac {\sin^{-1}x}{2}+C\\ \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.