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so I've just had my first exam (went pretty well) but I ran into this thing as the first part of the last question.

$$\int \frac{\cos x}{\sin x+\cos x}dx$$

I had a look on wolfram after the exam and it advised to multiply top and bottom by $\sec^3x$.

Is there another way to tackle this if you didn't know that trick? I find it hard to believe I was meant to know this and it was disproportionately harder than any type of integration question I've come across when practicing.

I couldn't get anywhere when trying to solve this.

Thanks.

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    $\begingroup$ If all else fails, the tangent half-angle substitution will handle this. If you have this as a definite integral over any of several well-behaved intervals, then it's easy. But the indefinite integral is more work. en.wikipedia.org/wiki/Tangent_half-angle_substitution $\endgroup$ – Michael Hardy Jan 20 '17 at 17:29
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Note that $\cos x = \frac{1}{2} ( 2 \cos x + \sin x - \sin x)$. So then $$ \int \frac{\cos x}{\sin x + \cos x} dx = \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x}dx + \frac{1}{2}\int \frac{\cos x + \sin x}{\cos x + \sin x} dx.$$ The second is easy to integrate, and the first is now just a $u$-substituteion as the numerator is the derivative of the denominator.

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  • $\begingroup$ Yes! That is the approach to use. (+1). $\endgroup$ – Mark Viola Jan 20 '17 at 17:39
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Hint: $\sin(x)+\cos(x)=\sqrt{2}\sin(x+\pi/4)$ Then substitute $u=x+\pi/4$.

A more general approach would be the following. Replace $\cos(x)$ and $\sin(x)$ by Euler's formula.

$$\int \frac{f(\cos(x),\sin(x))}{g(\cos(x),\sin(x))}dx=\int \frac{f(\exp(ix),\exp(-ix))}{g(\exp(ix),\exp(-ix))}dx.$$

Now, substitute $u=\exp(ix) \implies du=iudx$.

$$\int \frac{f(u,u^{-1})}{g(u,u^{-1})}\frac{du}{iu}=\int \frac{f(u)}{g(u)}\frac{du}{iu}.$$

If $f$ and $g$ are simple polynomials the resulting integral can always be solved by partial fractions.

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Method 1:-$$\frac{cosx}{cosx+sinx}=\frac{cos(x-π/4 +π/4)}{\sqrt {2}cos(x-π/4)}=\frac{cos(x-π/4)-sin(x-π/4)}{2cos(x-π/4)}$$

Method 2:- Write the numerator

$cosx=A(cosx+sinx)+B\frac{d}{dx}(cosx+sinx)$

to find $A$ and $B$ then the answer will be $Ax+Blog|cosx+sinx|+c$

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Dividing numerator and denominator by $\cos x$ yields

$$\frac{dx}{\tan x+1}$$

Now substitute $x=\tan^{-1}t,dx=\dfrac{dt}{1+t^2}$ to get

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

which can now be solved with partial fractions. Similar to the other method but hopefully easier to see.

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HINT:

For $I=\displaystyle\int\dfrac{a\cos x+b\sin x}{c\cos x+d\sin x}dx,$

write $a\cos x+b\sin x=A(c\cos x+d\sin x)+B\cdot\dfrac{d(c\cos x+d\sin x)}{dx}$ where $A,B$ are arbitrary constants so that

$I=\displaystyle A\int\ dx+B\int\dfrac{d(c\cos x+d\sin x)}{c\cos x+d\sin x}$

Compare the coefficients of $\cos x,\sin x$ to find $A,B$

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