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The integral I want to find is$$I=\int\frac{\sin{x}}{\sin{x}+\cos{x}}dx$$
The way I learnt is to introduce$$J=\int\frac{\cos{x}}{\sin{x}+\cos{x}}dx$$
Then $J+I=x+C_1$ and $J-I=\ln|\sin{x}+\cos{x}|+C_2$.
Is there some simple way to solve this integral $I$? For example, do not introduce other integrals?
Any hints will be appreciated. Thank you.

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\begin{align} \int\frac{\sin{x}}{\sin{x}+\cos{x}}dx&=\int\frac{\sin{x}}{\sqrt 2\sin{(x+\pi/4)}}dx\\ &=\int\frac{\sin{(u-\pi/4)}}{\sqrt 2\sin{(u)}}du\\ &=\frac12\int(1-\cot u)du\\ &=\frac u2-\frac12\log|\sin u|+c\\ &=\frac x2-\frac12\log|\sin x+\cos x|+c'\\ \end{align}

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  • $\begingroup$ A good and simple solution. Just a little question: is it compound angle formula from the second line to the third? thank you $\endgroup$ – Brian Cheung Jul 16 '15 at 9:15
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    $\begingroup$ From the 2nd to 3rd (and 4th to 5th), I used $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$. You are welcome! $\endgroup$ – Math-fun Jul 16 '15 at 10:13
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$$\int \frac{\sin x}{\sin x +\cos x}\,dx$$

$$=\frac{1}{2}\int \frac{(\sin x+\cos x)-(\cos x-\sin x)}{\sin x+\cos x}\,dx$$

$$=\frac{x}{2}-\frac{1}{2}\ln(\sin x+\cos x)+C$$

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    $\begingroup$ Well then it is actually the same as my solution $\endgroup$ – Brian Cheung Jul 16 '15 at 7:13
  • $\begingroup$ I think it is the simplest way...... $\endgroup$ – Empty Jul 16 '15 at 7:16
  • $\begingroup$ It is a simple way but I think it is hard to think of for beginners like me...... $\endgroup$ – Brian Cheung Jul 16 '15 at 7:19
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    $\begingroup$ If it is hard to think to you then I think putting $t=\tan(x/2)$ is too much hard to think.. $\endgroup$ – Empty Jul 16 '15 at 7:21
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The reasoning behind this trick: (or how to avoid rabbits pulled out of a hat)

$$\frac{\sin(x)}{\sin(x)+\cos(x)}$$ cannot be integrated by inspection, as it is lacking a $-\cos(x)$ term that would enable the use of

$$\frac{-\cos(x)+\sin(x)}{\ \ \ \sin(x)+\cos(x)}=-\frac{(\sin(x)+\cos(x))'}{\sin(x)+\cos(x)}.$$

It is tempting to add and remove this term, giving

$$\frac{-\cos(x)+\sin(x)}{\ \ \ \sin(x)+\cos(x)}+\frac{\cos(x)}{\sin(x)+\cos(x)}.$$

Then it suffices to observe that

$$\frac{\cos(x)}{\sin(x)+\cos(x)}=1-\frac{\sin(x)}{\sin(x)+\cos(x)},$$

and we are back to where we started from. This is good news as we have established

$$I=-\ln|\sin(x)+\cos(x)|+x-I.$$


For integrals of this type, the so-called Weierstrass substitutions $t=\tan(x/2)$ is a classical one.

Another is based on complex numbers,

$$\cos(x)=\frac{z+z^{-1}}2,\sin(x)=\frac{z-z^{-1}}{2i},dx=\frac{dz}{iz}.$$

Both will rationalize the integral, i.e. turn it to the ratio of two polynomials, for which a systematic method exists.

You should practice these.

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  • $\begingroup$ (+1) very nice answer. i like how you explain the logic behind this trick! Deserves many more upvotes! $\endgroup$ – tired Jul 16 '15 at 10:47
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hint: $t = \tan\left(\dfrac{x}{2}\right)$

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  • $\begingroup$ So should I use t-substitution then trigonometric substitution again to duel with the denominator? $\endgroup$ – Brian Cheung Jul 16 '15 at 7:21
  • $\begingroup$ Yes you should. $\endgroup$ – DeepSea Jul 16 '15 at 7:45
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HINT:

For $\int\dfrac{a\sin x+b\cos x}{A\sin x+B\cos x}dx$

write $\dfrac{a\sin x+b\cos x}{A\sin x+B\cos x}=C+D\dfrac{\dfrac{d(A\sin x+B\cos x)}{dx}}{A\sin x+B\cos x}$

$\implies a\sin x+b\cos x=C(A\sin x+B\cos x)+D(A\cos x-B\sin x)$

Compare the coefficients of $\sin x,\cos x$ to find $C,D$ in terms of $a,A,b,B$

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As the given $$I=\int \frac{sinx}{sinx+cosx}dx$$ $$= \int \frac{tanx}{tanx+1}dx$$ Put $ tanx=t \implies sec^2x.dx=dt$

$\implies (1+tan^2x)dx=dt \implies dx=\frac{dt}{1+t^2}$

So we have $$I=\int\frac{t}{t+1}.\frac{dt}{1+t^2} \to (1)$$ Now by using partial fraction $$\frac{t}{(1+t)(1+t^2)}=\frac{A}{1+t}+\frac{Bt+C}{1+t^2} \to (2) $$ $$t=A(1+t^2)+(Bt+C)(1+t) \to(3)$$ Put $1+t=0 \implies t=-1$ in above eq $$-1=A(1+1)+0 \implies A=\frac{-1}{2}$$ Now from eq $(3)$ $$t=A+At^2+Bt+Bt^2+C+Ct$$ Comparing coefficients of $t^2$ and $t$.

Coefficient of $t^2$ $$0=A+B \implies B=-A$$ $$B=\frac{1}{2}$$ Coefficient of $t$ $$1=B+C \implies C=1-B $$ $$C=1-\frac{1}{2} \implies C=\frac{1}{2} $$ Putting these values in eq $(2)$ $$\frac{t}{(1+t)(1+t^2)} = \frac{\frac{-1}{2}}{1+t} + \frac{\frac{1}{2}t+\frac{1}{2}}{1+t^2}$$ Putting this value in eq $(1)$ $$I=\int (\frac{\frac{-1}{2}}{1+t} + \frac{\frac{1}{2}t+\frac{1}{2}}{1+t^2})dt $$ $$I= \frac{1}{2} \int \frac{-1}{1+t}dt + \frac{1}{2} \int \frac{t+1}{t^2+1}dt $$ $$I= - \frac{1}{2} \ln (1+t) + \frac{1}{4}\int \frac{2t}{t^2+1}dt + \frac{1}{2} \int \frac{1}{t^2+1}dt$$ $$I= - \frac{1}{2} \ln (1+t) + \frac{1}{4} \ln (t^2+1) + \frac{1}{2} \arctan(t)+C $$ As $t=tanx$. So we have $$I= - \frac{1}{2} \ln (1+tanx) + \frac{1}{4} \ln (tan^2x+1) + \frac{1}{2} \arctan(tanx)+C $$ $$I= - \frac{1}{2} \ln (1+tanx) + \frac{1}{4} \ln (tan^2x+1) + \frac{1}{2} x +C$$ $$=-\frac{1}{2}\ln(1+tanx)-\frac{1}{2}\ln(cosx)+\frac{1}{2}x+C$$ $$=\frac{1}{2}x-\frac{1}{2}\ln(sinx+cosx)+C$$

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Since $$ \frac{\sin x}{\sin x+\cos x}=\frac{1}{2}\bigl(1+\tan(x-\pi/4)\bigr), $$ we get $$ \int \frac{\sin x}{\sin x+\cos x}\,dx=\frac{1}{2}\bigl(x+\ln|\cos(x-\pi/4)|\bigr)+C. $$

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