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$$I = \int_0^\frac{\pi}{2}g(x)\,dx = \int_0^\frac{\pi}{2}{\frac{\cos x}{\cos x + \sin x}}\,dx$$

This is part of a 3-part exercise where I was given $f, g:[0,\frac{\pi}{2}] \to \mathbb R \,\,\, $, $\,\,f(x)=\frac{\sin x}{\cos x+\sin x} \,\,\,$ , $\,\,g(x)=\frac{\cos x}{\sin x+\cos x} $

And these are the two things that I had to prove previously:

$(1)$ $\int_0^\frac{\pi}{2}\big({f(x)+g(x)\big)}dx = \frac{\pi}{2}$
$(2)$ $h:[0,\frac{\pi}{2}] \to \mathbb R\,\,\, $, $\,\,h(x)=\ln(\sin x + \cos x)$ is a primitive of $g-f$

I am sure that I need to use one or both of the relations above to solve this integral but I can't figure it out. I can see that $(1)$ is equivalent to $\int_0^\frac{\pi}{2}{g(x)} \, dx = \frac{\pi}{2} - \int_0^\frac{\pi}{2}f(x) \, dx $

I tried to solve the integral using u-sub and integration by parts and also tried to rewrite the integral using trigonometric identities but I had no succes so I am certain we need to make use of $(1)$ or/and $(2)$.

Michael Hardy
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Radu Gabriel
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4 Answers4

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Let $I = \int_0^{\frac{\pi}{2}}\frac{\cos(x)}{\sin(x)+\cos(x)}dx$.

Note that $\int_0^{\frac{\pi}{2}}\frac{\cos(x)-\sin(x)}{\sin(x)+\cos(x)}dx = [\ln(\sin(x)+\cos(x))]_0^{\frac{\pi}{2}}=0$.

Hence $I = \int_0^{\frac{\pi}{2}}\frac{\sin(x)}{\sin(x)+\cos(x)}dx$.

Next up,

$\int_0^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin(x)+\cos(x)}dx = \frac{\pi}{2}$.

Hence $2I = \frac{\pi}{2}$, and $I = \frac{\pi}{4}$

fGDu94
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    This certainly makes the problem more complicated than it needs to be, since there is no need to find any antiderivatives. – Michael Hardy Feb 12 '20 at 02:36
  • this is just what I saw when I looked at the problem, there is no strategy as such. – fGDu94 Feb 12 '20 at 02:37
  • And still it makes the problem appear more complicated than it is. – Michael Hardy Feb 12 '20 at 02:38
  • this is an absurdly simple proof, I wrote it in 30 seconds – fGDu94 Feb 12 '20 at 02:38
  • Nonetheless there's a simpler way. – Michael Hardy Feb 12 '20 at 02:40
  • @MichaelHardy. This is an exercise in a 12th grade high school book. It might seem harder to you to solve this way but I looked at your answer and that substitution is totally out of reach for anyone in the 12th grade. – Radu Gabriel Feb 12 '20 at 12:14
  • @RaduGabriel : It is nonsense to say my solution is out of reach to all 12th-graders. It is a substitution of the kind taught to 12th-graders. I did things like that when I was in 12th-grade. What typical 12th-graders do is that they're in it for grades or because someone pressured them to be there---both ignoble reasons; people like that don't count in this. – Michael Hardy Feb 12 '20 at 16:33
  • have to agree with Michael on this one, substitutions of this kind are a great tool – fGDu94 Feb 12 '20 at 16:38
  • @MichaelHardy. I understand why you think that. High school might be different in US, I guess you'll have to trust me that this answer is easier for me to understand. – Radu Gabriel Feb 13 '20 at 18:05
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This solution is the same as solutions above but with more details.

Start with $x\to \frac{\pi}{2}-x$ to have

$$I=\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}\ dx=-\int_{\pi/2}^0\frac{\cos(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)+\sin(\frac{\pi}{2}-x)}\ dx=\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\ dx$$

Add the integral to both sides,

$$\Longrightarrow I+I=\int_0^{\pi/2}\frac{\cos x+\sin x}{\cos x+\sin x}\ dx=\int_0^{\pi/2}dx=\frac{\pi}{2}\Longrightarrow I=\frac{\pi}{4}$$

Ali Shadhar
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\begin{align} & \frac{\cos x}{\sin x + \cos x} \, dx = \frac{\sin u}{ \cos u + \sin u} (-du) \\[12pt] \text{where } & u = \frac \pi 2 - x, \\[6pt] \text{ so that } & \cos x = \sin u \text{ and } \sin x = \cos u. \end{align} And as $x$ goes from $0$ to $\pi,$ $u$ goes from $\pi$ to $0.$

So the integrals of of $f$ and $g$ are equal to each other.

Michael Hardy
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You have found the primitive of $g - f$, and the primitive of $g + f = 1$ is obvious. Now write it as follows: $$ \int_0^\frac{\pi}{2} g(x) \; \mathrm{d}x = \frac{1}{2}\int_0^\frac{\pi}{2} (g(x) + f(x)) \; \mathrm{d}x + \frac{1}{2}\int_0^\frac{\pi}{2} (g(x) - f(x)) \; \mathrm{d}x $$

Clement Yung
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