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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)}$

Help please.

  • 0
    I came accros it before.2012-10-26

5 Answers 5

9

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 $I$

  • 0
    Specifically with what?2012-10-26
2

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.

  • 1
    Thanks for the link to Weierstrass substituion2012-10-26
2

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 ${1 \over 2}\int\frac{\cos(a + {\pi \over 4}) + \sin(a + {\pi \over 4})}{\sin(a + {\pi \over 4})}\,da$ ${1 \over 2}\int (\cot(a + {\pi \over 4}) + 1) \,da$ $= {1 \over 2}\ln|\sin(a + {\pi \over 4})| + {a \over 2} + C$

1

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.

  • 0
    @MaoYiyi, Type-V and Type-VI of http://www.me.berkeley.edu/~ndhillon/Teaching/Maths/bookch4.html.2012-10-26
1

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