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Today I passed the test to access to medicine, and I found some really cool questions on the mathematics part, anyways I just went to see the difficulty of the test, and sorry for being far from my main question : $$\int_{\pi/6}^{\pi/3} \frac{\sin x}{\sin x+\cos x} \mathrm{d}x$$ I used (obviously) the king property of integration which gives $\pi/12$ as an answer.

My question is how can we solve this using some basic properties of integration ?

Thanks in advance. and please wish me some luck for the next tests to access engineering universities.

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3 Answers 3

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HINT

Make the change of variable $y = \pi/2 - x$. Then we get that \begin{align*} I & = \int_{\pi/6}^{\pi/3}\frac{\sin(x)}{\sin(x) + \cos(x)}\mathrm{d}x = \int_{\pi/6}^{\pi/3}\frac{\cos(y)}{\cos(y) + \sin(y)}\mathrm{d}y \end{align*}

Then add both expressions together and see what happens.

Can you take it from there?

EDIT

To begin with, notice that \begin{align*} \frac{\sin(x)}{\sin(x) + \cos(x)} & = \frac{1}{2}\left(\frac{2\sin(x)}{\sin(x) + \cos(x)}\right)\\\\ & = \frac{1}{2}\left[\frac{(\sin(x) + \cos(x)) + (\sin(x) - \cos(x))}{\sin(x) + \cos(x)}\right]\\\\ & = \frac{1}{2} - \frac{1}{2}\left(\frac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)}\right) \end{align*}

Moreover, we do also have that $(\sin(x) + \cos(x))' = \cos(x) - \sin(x)$.

Having said that, according to the substitution $u = \sin(x) + \cos(x)$, we get that \begin{align*} \int\frac{\sin(x)}{\sin(x) + \cos(x)}\mathrm{d}x = \frac{x}{2} - \frac{\ln(\sin(x) + \cos(x))}{2} + c \end{align*}

Now it remains to apply the integration limits.

Hopefully this is helpful!

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  • $\begingroup$ I literally used this technique is there any metjod to evaluate it using anti derivatives or integration by parts.. ? $\endgroup$ Jul 15, 2021 at 21:29
  • $\begingroup$ Yes, there is. I have provided it in the edit. $\endgroup$
    – user0102
    Jul 15, 2021 at 21:40
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If $S$ is your integral and$$C=\int_{\pi/6}^{\pi/3}\frac{\cos x}{\sin x+\cos x}\,\mathrm dx,$$then $S+C=\frac\pi6$. On the other hand\begin{align}C&=\int_{\pi/6}^{\pi/3}\frac{\cos x}{\sin x+\cos x}\,\mathrm dx\\&=-\int_{\pi/2-\pi/6}^{\pi/2-\pi/3}\frac{\cos\left(\frac\pi2-x\right)}{\sin\left(\frac\pi2-x\right)+\cos\left(\frac\pi2-x\right)}\,\mathrm dx\\&=-\int_{\pi/3}^{\pi/6}\frac{\sin x}{\cos x+\sin x}\,\mathrm dx\\&=\int_{\pi/6}^{\pi/3}\frac{\sin x}{\cos x+\sin x}\,\mathrm dx\\&=S.\end{align}So, since $S+C=\frac\pi6$ and $S=C$, $S=\frac\pi{12}$.


Another possibility is this: note that $\sin(x)+\cos(x)=\sqrt2\sin\left(x+\frac\pi4\right)$ and that $\cos(x)=\cos\left(x+\frac\pi4\right)\frac1{\sqrt2}+\sin\left(x+\frac\pi4\right)\frac1{\sqrt2}$. So\begin{align}\int_{\pi/6}^{\pi/3}\frac{\cos x}{\sin x+\cos x}\,\mathrm dx&=\frac12\int_{\pi/6}^{\pi/3}\cot\left(x+\frac\pi4\right)+1\,\mathrm dx\\&=\frac12\left(\left(\log\left(\sin\left(\frac{7\pi}{12}\right)\right)+\frac\pi3\right)-\left(\log\left(\sin\left(\frac{5\pi}{12}\right)\right)+\frac\pi6\right)\right)\\&=\frac\pi{12},\end{align}since $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{7\pi}{12}\right)$.

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  • $\begingroup$ Very nice all the answers for this question. $\endgroup$
    – Sebastiano
    Jul 15, 2021 at 22:15
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Substituting $t=\tan\left(\frac x2\right)$ is a well-known approach to this kind of integral. Employing it yields the integral of a fairly simple rational function,

$$\begin{align} \int_{\pi/6}^{\pi/3}\frac{\sin(x)}{\sin(x)+\cos(x)}\,\mathrm dx &= \int_{2-\sqrt3}^{1/\sqrt3}\frac{\frac{2t}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}\cdot\frac{2\,\mathrm dt}{1+t^2} \\[1ex] &= \int_{2-\sqrt3}^{1/\sqrt3}\frac{4t}{(1+2t-t^2)(1+t^2)}\,\mathrm dt \end{align}$$

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