# What is $\int\frac{dx}{x+\sqrt{1-x^2}}$? [duplicate]

Possible Duplicate:
Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$

$$\int\frac{dx}{x+\sqrt{1-x^2}}$$

I used $x = \sin(a),\ dx = \cos (a)\,da$.

However, I am getting suck after that.

I can't seem to find what to do after $$\int \frac{\cos(a)\,da}{\sin(a)+\cos(a)}$$

## marked as duplicate by Gerry Myerson, N. S., Hans Lundmark, Noah Snyder, rschwiebOct 26 '12 at 12:37

• – A. Chu Oct 26 '12 at 4:12
• @jasoncube thanks, how did you find that? – yiyi Oct 26 '12 at 4:24
• I came accros it before. – A. Chu Oct 26 '12 at 6:33

Try this: $$I=\int \frac{\cos x dx}{\cos x+ \sin x}\\ J=\int \frac{\sin x dx}{\cos x+ \sin x}$$ Then, $$I+J=x+C_1\\ I-J=\int \frac{(\cos x - \sin x )dx}{\cos x+ \sin x}=\int d \log(\cos x + \sin x)\\ =\log(\cos x + \sin x) +C_2$$ Hence you have two unknowns in two equations, from which you can easily find $I$

• Could you explain how to came up with this. – yiyi Oct 26 '12 at 3:51
• Specifically with what? – Alex Oct 26 '12 at 4:05

If you don't see the trick, you can try using the formula for $A \cos(t) + B \sin(t)$, namely $\sin(a) + \cos(a) = \sqrt{2}(\sin(a + {\pi \over 4}))$. So you seek $${1 \over \sqrt2}\int \frac{\cos(a)}{\sin(a + {\pi \over 4})}\,da$$ $$= {1 \over \sqrt2}\int \frac{\cos(a + {\pi \over 4} - {\pi \over 4})}{\sin(a + {\pi \over 4})}\,da$$ Using the cosine subtraction formula, this is $${1 \over 2}\int\frac{\cos(a + {\pi \over 4}) + \sin(a + {\pi \over 4})}{\sin(a + {\pi \over 4})}\,da$$ $${1 \over 2}\int (\cot(a + {\pi \over 4}) + 1) \,da$$ $$= {1 \over 2}\ln|\sin(a + {\pi \over 4})| + {a \over 2} + C$$

Hint or starting point:

Nothing immediate jumps out to me when I see the integral

$$\int \frac{\cos \theta}{\sin \theta + \cos\theta} d\theta$$

However, when integrating rational function of trig functions, we can always use the Weierstrass substitution.

So, let $t = \tan \left(\frac{\theta}{2}\right), dt = \frac{1}{2}\sec^2\left(\frac{\theta}{2}\right) \ dx.$ With the Weierstrass sub, we have that: $$\sin \theta = \frac{2t}{1+t^2}, \cos\theta = \frac{1-t^2}{1+t^2}, d\theta = \frac{2 \ dt}{1+t^2}$$

This allows us to rewrite our integral as:

$$\int\frac{2(1-t^2) \ dt}{(t^2+1)^2 \left(\frac{2t}{t^2+1} + \frac{1-t^2}{t^2+1}\right) }$$

Clean up the denominator to get:

$$2\int\frac{(t^2-1) \ dt}{t^4 - 2t^3 - 2t - 1}$$

From here, looks like partial fractions will do it. Note the denominator factors as

## $$(t^2+1)(t^2-2t-1)$$

Second thought Write the original integral as $$\int \frac{1}{1+\cot \theta} d\theta$$ Let $u = \cot \theta, du = -(1+u^2) d\theta$. Then we have:

$$-\int\frac{1}{(1+u^2)(u^2+1)} \ du$$

This is a nicer partial fraction decomposition to work with and is less work in getting there compared to my original thinking.

• Thanks for the link to Weierstrass substituion – yiyi Oct 26 '12 at 3:52

The substitution $x=2t/(1+t^2)$ will reduce the original integrand to a rational (though perhaps messy) function of $t$.

Put $\cos a=A(\sin a+\cos a)+B\frac{d(\sin a+\cos a)}{da}$ where $A,B$ are constants

Now, $\cos a=A(\sin a+\cos a)+B(\cos a-\sin a)$

Equating the coefficients of $\sin a,A-B=0$ or $A=B$

Equating the coefficients of $\cos a,A+B=1$ or $A=B=\frac 1 2$

So, $\int \frac{\cos a da}{\sin a+\cos a}$ $$=\frac 1 2\int da+\frac 1 2\int\frac{d(\sin a+\cos a)}{\sin a+\cos a}=\frac a 2+\frac 1 2\log(\sin a+\cos a)+C$$ where $C$ is the indeterminate constant of indefinite integral.

• How did you come up with that subitution? – yiyi Oct 26 '12 at 3:51
• @MaoYiyi, this is the approach for $\frac{A\sin x+B\cos x+C}{D\sin x+E\cos x+F}$ – lab bhattacharjee Oct 26 '12 at 3:56
• Do you have the page numbers of a book, where I can read up on that. – yiyi Oct 26 '12 at 3:57
• @MaoYiyi, Type-V and Type-VI of me.berkeley.edu/~ndhillon/Teaching/Maths/bookch4.html. – lab bhattacharjee Oct 26 '12 at 4:02