Expert: Paul Klarreich - 5/7/2007

Question
Hello Sir

I am having some problem solvong this integral. Looks simple, but i cant get it right; please help.
For an angle 0<q<90 , integral from -q to q of x/cosx = 0.
Frm this how do i deducethe integral from 0 to 2q of x/cos(x-q) ??????//

Answer
Questioner:   mohammad
Category:  Calculus
 
Subject:  Fractional Integral
Question:  Hello Sir

I am having some problem solvong this integral. Looks simple, but i cant get it right; please

help.
For an angle 0<q<90 , integral from -q to q of x/cosx = 0.
Frm this how do i deducethe integral from 0 to 2q of x/cos(x-q) ??????//

.............................
Hi, Mohammed,

Basic rule for 'dummy variables':  All of these are the same:

{q      x
|     ----- dx
}-q   cos x

{q      y
|     ----- dy
}-q   cos y

{q      t
|     ----- dt
}-q   cos t

{q      w
|     ----- dw
}-q   cos w

That is, the variables x,y,t,w are all dummy variables.  That means they don't appear in the final

answer -- they are just place-holders.  The answer may involve q, but not any of those.

{q      x
|     ----- dx = 0, which is given.
}-q   cos x

Your task is to compute:
 
{2q        x
|     ----------- dx
}0    cos (x - q)

Make a change of variable:

t = x - q
x = t + q
dx = dt

x = 0  -->  t = -q
x = 2q -->  t = q

Your integral becomes:

{q     t + q
|     ------- dt =
}-q    cos t

{q       q
|     ------- dt
}-q    cos t

Now that last step was real tricky.  Here's what I did:

{q     t + q
|     ------- dt =
}-q    cos t

{q       t
|     ------- dt   PLUS
}-q    cos t

{q       q
|     ------- dt
}-q    cos t

But the first term is exactly (see that discussion of dummy variables) zero, because it is exactly

the given integral.

So this is all we have to evaluate:

{q       q
|     ------- dt
}-q    cos t

Rewrite as:

{q
|    q sec t dt
}-q

Now look up a standard integral:  (see note at the bottom)

{
| sec t dt = ln | sec t + tan t }
}

Evaluate it from  t = -q to t = q:

q [ ln | sec q + tan q | - ln | sec(-q) + tan(-q) | ]

Now use basic odd-even properties:

sec(-q) = sec q,  since  secant is an even function.
tan(-q) = - tan q, since  tangent is odd.

You have:

q [ ln | sec q + tan q | - ln | sec q - tan q | ] =

       sec q + tan q
q ln | ---------------- |
       sec q - tan q

I think that does it.  Well, sort of.  You might see a different answer in the back of the book,

so here is a different way:

************* ALTERNATIVE APPROACH ****************
So this is all we have to evaluate:

{q       q
|     ------- dt
}-q    cos t

Now cosine is an even function, and a basic property of even functions says:

{q   
|    evenfunction(x) dx =
}-q  

{q   
|   2 evenfunction(x) dx =
}0  

So we can rewrite as:

{q
|   2q sec t dt
}0

Now look up a standard integral:

{
| sec t dt = ln | sec t + tan t }
}

Evaluate it from  t = 0 to t = q:

2q [ ln | sec q + tan q | - ln | sec(0) + tan(0) | ]

2q [ ln | sec q + tan q | - ln | 1 | ]

2q [ ln | sec q + tan q | - 0 ]

2q ln | sec q + tan q |

Maybe that does it.

.....................................
Note at the bottom:

If you don't have the table of integrals, or are not permitted to use it, you can prove that:

{
| sec t dt = ln | sec t + tan t }
}

by doing this trick:

{
| sec t dt =
}
{ sec t(sec t + tan t)
| -------------------- dt
}   sec t + tan t

We just multiplied top and bottom by  sec t + tan t.  Now the top, after you multiply out, is exactly the derivative of the bottom.