Solution to STEP problem-I

I have solved a STEP problem from Stephen Siklos’ book Advanced Problems in Core Mathematics. The problem and the solution follows:

Problem: Use the substitution x=2-\cos\theta to evaluate

\int_{3/2}^{2} (\frac {x-1}{3-x})^{1/2}\, dx

Hence or otherwise show that for a<b

\int_{p}^{q} (\frac {x-a}{b-x})^{1/2}\, dx=\frac {(b-a)(\pi+3\sqrt{3}-6)}{12}

Where p=\frac {3a+b}{4} and q=\frac {a+b}{2}

Solution:

At the first look it seems to me that solving the first part may give an idea of the second part. So lets begin with the first part of the question.

\int_{3/2}^{2} (\frac {x-1}{3-x})^{1/2}\, dx

Lets substitute x with 2-\cos \theta

As x=2-\cos \theta

\Rightarrow dx=\sin \theta d\theta

Now lets deal with the limits:

for x=\frac {3}{2}

becomes \frac {3}{2}=2-\cos \theta

\Rightarrow \cos \theta=2-\frac {3}{2}

\Rightarrow \cos \theta=\frac {1}{2}

\Rightarrow \theta=\cos^{-1} \frac {1}{2}

\therefore \theta=\frac {\pi}{3}

Similarly for x=2 is 2=2-2\cos \theta

\Rightarrow \cos \theta=0

\Rightarrow \theta=\cos^{-1}0

\therefore \theta=\frac {\pi}{2}

Now the integral becomes:

\int_{\pi/3}^{\pi/2} (\frac {1-\cos \theta}{1+\cos \theta})^{1/2}\, \sin \theta d\theta

=\int_{\pi/3}^{\pi/2} \tan {\frac {\theta}{2}} \sin \theta\, d\theta

=\int_{\pi/3}^{\pi/2} \tan {\frac {\theta}{2}} 2\sin {\frac {\theta}{2}}\cos{\frac {\theta}{2}}\, d\theta

=\int_{\pi/3}^{\pi/2} 2\sin^{2} {\frac {\theta}{2}}\, d\theta

=\int_{\pi/3}^{\pi/2} (1-\cos \theta)\, d\theta

=\int_{\pi/3}^{\pi/2} d\theta-\int_{\pi/3}^{\pi/2} \cos \theta d\theta

=\left [\frac {\pi}{2}-\frac {\pi}{3} \right]-[\sin \frac {\pi}{2}-\sin \frac {\pi}{3}]

=\frac {\pi}{6}-\left (1-\frac {\sqrt 3}{2} \right )

=\frac {\pi}{6}+\frac {\sqrt 3}{2}-1

=\frac {\pi+3\sqrt 3-6}{6}

=\frac {2(\pi+3\sqrt 3-6)}{6}

=\frac {(3-1)(\pi+3\sqrt 3-6)}{6}

Now, we notice that this is a special case of the equation:

\int_{p}^{q} (\frac {x-a}{b-x})^{1/2}\, dx=\frac {(b-a)(\pi+3\sqrt{3}-6)}{12} ; a<b

Where p=\frac {3a+b}{4} and q=\frac {a+b}{2}

Now we turn to the second part of the problem. I think if we could express the fraction \frac {x-a}{b-x} as \frac {1-\cos \theta}{1+\cos \theta} then we would find the solution.In order to do so we should find a number r for which r-a=b-r

So r=\frac {a+b}{2}

And \frac {a+b}{2}-a=b-\frac {a+b}{2}=\frac {b-a}{2}

So we should replace x with \frac {a+b}{2}-\frac {b-a}{2} \cos \theta

As x=\frac {a+b}{2} -\frac {b-a}{2} \cos \theta

So dx=\frac {b-a}{2} \sin \theta d\theta

Now lets evaluate the limits for x=\frac {a+b}{2} -\frac {b-a}{2} \cos \theta

For p=\frac {3a+b}{4} is

\frac {3a+b}{4}=\frac {a+b}{2} -\frac {b-a}{2} \cos \theta

\Rightarrow \frac {b-a}{2} \cos \theta=\frac {a+b}{2}-\frac {3a+b}{4}

\Rightarrow \frac {b-a}{2} \cos \theta=\frac {2a+2b-3a-b}{4}

\Rightarrow \frac {b-a}{2} \cos \theta=\frac {b-a}{4}

\Rightarrow \cos \theta=\frac {1}{2}

\Rightarrow \theta=\cos^{-1} \frac {1}{2}

\therefore \theta=\frac {\pi}{3}

Similarly for q=\frac {a+b}{2}

\frac {a+b}{2}=\frac {a+b}{2}-\frac {b-a}{2}\cos \theta

\Rightarrow \frac {b-a}{2}\cos \theta=0

\Rightarrow \cos \theta=0

\theta=\cos^{-1} 0

\therefore \theta=\frac {\pi}{2}

Now the second integral becomes:

\int_{\pi/3}^{\pi/2} (\frac {\frac {a+b}{2}-\frac {b-a}{2} \cos \theta-a}{b-\frac {a+b}{2}+\frac {b-a}{2} \cos \theta})^{1/2} \frac {b-a}{2} \sin \theta d\theta

=\int_{\pi/3}^{\pi/2} (\frac {\frac {b-a}{2}-\frac {b-a}{2} \cos \theta}{\frac {b-a}{2}+\frac {b-a}{2} \cos \theta})^{1/2} \frac {b-a}{2} \sin \theta d\theta

=\frac {b-a}{2} \int_{\pi/3}^{\pi/2} (\frac {\frac {b-a}{2} (1-\cos \theta)}{\frac {b-a}{2} (1+\cos \theta)})^{1/2} \sin \theta d\theta

=\frac {b-a}{2} \int_{\pi/3}^{\pi/2} (\frac {1-cos \theta}{1+cos \theta})^{1/2} \sin \theta d\theta

=\frac {b-a}{2} \int_{\pi/3}^{\pi/2} \tan \frac {\theta}{2} \sin \theta d\theta

=\frac {b-a}{2} \int_{\pi/3}^{\pi/2} \tan \frac {\theta}{2} 2\sin \frac {\theta}{2}\cos \frac {\theta}{2} d\theta

=\frac {b-a}{2} \int_{\pi/3}^{\pi/2} 2 \sin^{2} \frac {\theta}{2} d\theta

=\frac {b-a}{2} \int_{\pi/3}^{\pi/2} (1-\cos \theta) d\theta

=\frac {b-a}{2} (\int_{\pi/3}^{\pi/2} d\theta-\int_{\pi/3}^{\pi/2} cos \theta d\theta)

=\frac {b-a}{2} ([\frac {\pi}{2}-\frac {\pi}{3}]-[\sin \frac {\pi}{2}-\sin \frac {\pi}{3}])

=\frac {b-a}{2} (\frac {\pi}{6}-[1-\frac {\sqrt 3}{2}])

=\frac {b-a}{2} (\frac {\pi}{6}+\frac {\sqrt 3}{2}-1)

=\frac {b-a}{2} (\frac {\pi+3\sqrt 3-6}{6})

=\frac {(b-a)(\pi+3\sqrt 3-6)}{12})

And we solve the problem!

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About f.nasim

A CS undergrad from Bangladesh.
This entry was posted in calculus, mathematics and tagged , , . Bookmark the permalink.

One Response to Solution to STEP problem-I

  1. Terry Tao says:

    Yes, your solution is correct. no problems i guess.

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