IMO Integration Problem: Evaluate $\displaystyle\int^{\dfrac{\pi}{4}}_0 \dfrac{x}{(\sin x+\cos x)\cos x}\ dx$.

Evaluate $\displaystyle\int^{\dfrac{\pi}{4}}_0 \dfrac{x}{(\sin x+\cos x)\cos x}\ dx$.

Solving for this problem is a breeze if you're someone who is so sensitive about the possibility of the existence of property of symmetry for the integrand function involved.

And note that this is a solution provided by an Indian friend of mine.

First, note that we could simplify the sum $\sin x+\cos x$ as:

$\begin{align*}\sin x+\cos x&=\cos x+\sin x\\&=\sqrt{1^2+1^2}\cos (x+\arctan 1)\\&=\sqrt{2}\cos \left(x+\dfrac{\pi}{4}\right)\end{align*}$

Therefore, we're dealing with the following definite integral:

$\displaystyle\int^{\dfrac{\pi}{4}}_0 \dfrac{x}{(\sin x+\cos x)\cos x}\ dx=\displaystyle\int^{\dfrac{\pi}{4}}_0\dfrac{x}{\sqrt{2}\cos \left(x+\dfrac{\pi}{4}\right)\cos x}\ dx$.

Now, if we let

[MATH]\begin{align*}I&=\int_0^{\pi/4} \frac{x}{(\sin x+\cos x)\cos x}\,dx\\&=\int_0^{\pi/4} \frac{x}{\sqrt{2}\cos \left(\frac{\pi}{4}-x\right)\cos x}\,dx\,\,(1)\end{align*}[/MATH]

And note that if we define $f(x)=\dfrac{x}{\sqrt{2}\cos \left(x+\dfrac{\pi}{4}\right)\cos x}$ and we then replace $x$ by $\dfrac{\pi}{4}-x$ into $f(x)$, we see that we have:

$\begin{align*}f\left(x+\dfrac{\pi}{4}\right)&=\dfrac{\left(x+\dfrac{\pi}{4}\right)}{\sqrt{2}\cos \left(\dfrac{\pi}{4}-\left(x+\dfrac{\pi}{4}\right)\right)\cos \left(x+\dfrac{\pi}{4}\right)}\\&=\dfrac{\left(x+\dfrac{\pi}{4}\right)}{\sqrt{2}\cos (-x)\cos \left(x+\dfrac{\pi}{4}\right)}\\&=\dfrac{\left(x+\dfrac{\pi}{4}\right)}{\sqrt{2}\cos (x)\cos \left(x+\dfrac{\pi}{4}\right)}\end{align*}$

which means $f(x)$ is symmetric with respect to the vertical line $x=\dfrac{\pi}{4}$ since $f\left(x+\dfrac{\pi}{4}\right)=\dfrac{\left(x+\dfrac{\pi}{4}\right)}{\sqrt{2}\cos (x)\cos \left(x+\dfrac{\pi}{4}\right)}$, is even.

Therefore, we also have

[MATH]I=\int_0^{\pi/4} \frac{\dfrac{\pi}{4}-x}{\sqrt{2}\cos \left(\dfrac{\pi}{4}-x\right)\cos x}\,dx\,\,\,(2)[/MATH]

Add $(1)$ and $(2)$ we get:

[MATH]I+I=\int_0^{\pi/4} \dfrac{x}{\sqrt{2}\cos \left(\dfrac{\pi}{4}-x\right)\cos x}\,dx+\int_0^{\pi/4} \dfrac{\frac{\pi}{4}-x}{\sqrt{2}\cos \left(\dfrac{\pi}{4}-x\right)\cos x}[/MATH]

[MATH]2I=\int_0^{\pi/4} \dfrac{x+\dfrac{\pi}{4}-x}{\sqrt{2}\cos \left(\dfrac{\pi}{4}-x\right)\cos x}\,dx[/MATH]

[MATH]2I=\int_0^{\pi/4} \dfrac{\cancel{x}+\dfrac{\pi}{4}-\cancel{x}}{\sqrt{2}\cos \left(\dfrac{\pi}{4}-x\right)\cos x}\,dx[/MATH]

[MATH]2I=\dfrac{\pi}{4\sqrt{2}}\int_0^{\pi/4} \dfrac{1}{\cos \left(\dfrac{\pi}{4}-x\right)\cos x}\,dx[/MATH]

Now comes the magic trick!

We could do the algebraic manipulation to make the finding of the anti-derivative easier:

[MATH]2I=\frac{\pi}{4\sqrt{2}}\cdot \frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{2}\int_0^{\pi/4} \dfrac{dx}{\cos \left(\dfrac{\pi}{4}-x\right)\cos x}[/MATH]

[MATH]2I=\frac{\pi}{4} \int_0^{\pi/4} \dfrac{\frac{\sqrt{2}}{2}\, dx}{\cos \left(\dfrac{\pi}{4}-x\right)\cos x}[/MATH]

[MATH]2I=\frac{\pi}{4} \int_0^{\pi/4} \dfrac{\sin \left(\dfrac{\pi}{4}-x+x\right)\, dx}{\cos \left(\dfrac{\pi}{4}-x\right)\cos x}[/MATH]

[MATH]I=\frac{\pi}{8} \int_0^{\pi/4} \dfrac{\sin \left(\dfrac{\pi}{4}-x\right)\cos x+\cos \left(\dfrac{\pi}{4}-x\right)\sin x}{\cos \left(\dfrac{\pi}{4}-x\right)\cos x}\, dx[/MATH]

[MATH]I=\frac{\pi}{8} \int_0^{\pi/4} \left(\dfrac{\sin \left(\dfrac{\pi}{4}-x\right)\cos x\, dx}{\cos \left(\dfrac{\pi}{4}-x\right)\cos x}+\dfrac{\cos \left(\dfrac{\pi}{4}-x\right)\sin x}{\cos \left(\dfrac{\pi}{4}-x\right)\cos x}\right)\,\, dx[/MATH]

[MATH]I=\frac{\pi}{8} \int_0^{\pi/4} \left(\tan \left(\dfrac{\pi}{4}-x\right)+\tan x \right),\, dx[/MATH]

But notice again [MATH] \int_0^{\pi/4} \tan \left(\dfrac{\pi}{4}-x\right),\, dx= \int_0^{\pi/4} \tan x,\, dx[/MATH]

Hence,

$\begin{align*}I&=\frac{\pi}{8}\int_0^{\pi/4} (2\tan x)\,dx\\&=\frac{\pi}{4}\int_0^{\pi/4} \tan x\,dx\\&=\frac{\pi}{4}\left(\ln(\sec x)\right|_0^{\pi/4}\\&=\frac{\pi}{4}\left(\ln\left(\sec \dfrac{\pi}{4}\right)-0\right)\\&=\frac{\pi}{4}\ln \sqrt{2}\\&=\frac{1}{2}\frac{\pi}{4}\ln 2\\&=\frac{\pi}{8} \ln 2\end{align*}$