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Thursday, August 6, 2015

IMO Integration Problem: Evaluate π40x(sinx+cosx)cosx dx.

Evaluate π40x(sinx+cosx)cosx 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 sinx+cosx as:

sinx+cosx=cosx+sinx=12+12cos(x+arctan1)=2cos(x+π4)

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

π40x(sinx+cosx)cosx dx=π40x2cos(x+π4)cosx dx.

Now, if we let

I=π/40x(sinx+cosx)cosxdx=π/40x2cos(π4x)cosxdx(1)

And note that if we define f(x)=x2cos(x+π4)cosx and we then replace x by π4x into f(x), we see that we have:

f(x+π4)=(x+π4)2cos(π4(x+π4))cos(x+π4)=(x+π4)2cos(x)cos(x+π4)=(x+π4)2cos(x)cos(x+π4)

which means f(x) is symmetric with respect to the vertical line x=π4 since f(x+π4)=(x+π4)2cos(x)cos(x+π4), is even.

Therefore, we also have

I=π/40π4x2cos(π4x)cosxdx(2)

Add (1) and (2) we get:

I+I=π/40x2cos(π4x)cosxdx+π/40π4x2cos(π4x)cosx

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

\displaystyle 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

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

Now comes the magic trick!

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

\displaystyle 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}

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

\displaystyle 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}

\displaystyle 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

\displaystyle 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

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

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



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*}


3 comments:

  1. Hi, the identity must be √cos(x-π4)=cos(x)+sin(x) with a negative sign in the cos.

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  2. This comment has been removed by the author.

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  3. multiply cosx in to the bracket then convert x in to 2x use king rule then take tan(x)=t

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