# Definite Integral (Calculus)

this is a revision problem, not a homework problem. Sincere thanks for help.

Question: Evaluate $\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}$.

The answer is $\pi/4$, but I am unable to work out the method.

Sincere thanks for help.

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Notice that $$\frac{d}{dx} (\log(\cos x + \sin x)) = \frac{\cos x - \sin x}{\cos x + \sin x}.$$ Now if $$I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx= \int_0^{\pi/2} 1 - \frac{\sin x}{\cos x + \sin x}dx$$ then we see that $$2I = \int_0^{\pi/2} 1 + \frac{\cos x - \sin x}{\cos x + \sin x} dx = \pi/2+0.$$

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Thanks, this method is natural and elegant. – yoyostein Jun 16 '12 at 10:01

Let $f(x)=\dfrac{\cos x}{\cos x+\sin x}$ and $g(x)=\dfrac{\sin x}{\cos x+\sin x}$. Note that $g\left(\frac{\pi}{2}-x\right)=f(x)$, which implies that $\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}g(x)dx$. Also, $f(x)+g(x)=1$ for all $x$, so $\int_0^{\pi/2}(f(x)+g(x))dx=\pi/2$.

Consequently,

$\frac{\pi}{2}=\int_0^{\pi/2}(f(x)+g(x))dx=\int_0^{\pi/2}f(x)dx+\int_0^{\pi/2}g(x)dx=2\int_0^{\pi/2}f(x)dx.$

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Community wiki? – The Chaz 2.0 Jun 16 '12 at 19:22
Phrase with a question mark at the end? – Jonas Meyer Jun 17 '12 at 15:49
Next time I'll be more explicit [interrobang] ¿¡ – The Chaz 2.0 Jun 17 '12 at 22:29
¿¡Sounds good!? – Jonas Meyer Jun 18 '12 at 4:08

Another easy way would be to directly write that:

$$\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}=\int_0^{\pi/2}\frac{1}{2}\left(1+\frac{\cos x - \sin x}{\cos x+\sin x}\right)=\frac{1}{2}(x+\ln(\sin x +\cos x))\big|_0^{\pi/2} = \frac{\pi}{4}.$$

Q.E.D.

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$$I = \int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}dx \tag{1}$$

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

Add (1) and (2) to get

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

Thus, $I$ = $\pi/4$

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Let $I:= \displaystyle{ \int_{0}^{\pi /2} \frac{\cos x}{\cos x+ \sin x } dx}$ and $J:= \displaystyle{ \int_{0}^{\pi /2} \frac{\sin x}{\cos x+ \sin x } dx}$.

Now we can see that: $I+J =\frac{\pi}{2}$ and $\displaystyle{I-J= \int_{0}^{\pi /2} \frac{ \cos x - \sin x}{\cos x + \sin x} dx = \log ( \sin x +\cos x) |_0^{\pi /2} } =0$

Solving this algebraic system we get that $\boxed{ I=J= \frac{\pi}{4}}$.

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