2

Let $x\in[0,\pi/2]$, $I(x)=\int_0^x \frac{\cos(t)}{\sin(t)+\cos(t)} dt$ and $J(x)=\int_0^x \frac{\sin(t)}{\sin(t)+\cos(t)} dt$ .

The aim of the exercise is to calculate $I(x)$ and $J(x)$.

Fisrt of all, we see that $I(x)+J(x)=x$.

If I do "$u=\pi/2 -t$", I get (since $\sin(\pi/2-y) = \cos(y)$): $I(x) = \int_{\pi/2-x}^{\pi/2} \frac{\sin(u)}{\cos(u)+\sin(u)}du = J(\pi/2) - J(x)$. And so $I(x)+J(x)= J(\pi/2)$... Which is impossible beacause $J(\pi/2)$ doesn't depend of $x$.

I don't see where my computations fails...

Sebastien
  • 834

1 Answers1

1

You have that

$$I(x) + J(x) = x \tag{1}$$

but note also

$$I(x) - J(x) = \int_0^x \frac{\cos t - \sin t}{\cos t + \sin t} dt. \tag{2}$$

This integral is of the form $$\int_0^x \frac{f'(t)}{f(t)} dt$$ so it evaluates to $$[\ln(|\cos t + \sin t|)]_0^x = \ln(|\cos x + \sin x|).$$

We can thus add the simultaneous equations $(1)$ and $(2)$ and divide through by $2$ to get

$$I(x) = \frac{x + \ln(|\cos x + \sin x|)}{2}$$

and hence also find that

$$J(x) = \frac{x - \ln(|\cos x + \sin x|)}{2}.$$

Robin
  • 6,201