Expert: Paul Klarreich - 11/24/2005

Question
(  ..e^(1/(x+1)
 ..----------   dx
) ..(x + 1)^2

(  ..sin^2 x-cos^2 x
  ..---------------  dx
)       ..sinx


(5e  .. 1
    ..--- dx
)1  .. x

(   .. x + 3
  ..--------  dx
)  ..x^2 + 9


I have to know how to do integral problems like these.  My test will me almost exactly like these and it is monday.  Thank you for your time and knowledge.


Answer
Hi, Jessica,

I have to know how to do integral problems like these. My test will me almost exactly like these and it is monday.

Never leave anything for the last minute, now.

( ..e^(1/(x+1)
..---------- dx
) ..(x + 1)^2

This is a 'basic' substitution.  When you look at this, you try to find some element whose derivative is a factor of the integrand.  

There are several 'styles' for handling this type of problem.  (Not different methods; they are all the same method.)  My recommended style is this:

Make a little 'box' in which you do your substitution work:

1. Let u = something.
2. Write  du/dx = derivative of that something.
3. Solve algebraically for  dx.

Now go back to the integral and substitute 'u' where you see it, and replace dx by whatever you got.

In this case, your substitution is:
========= BEGIN SUBSTITUTION BOX ===========
u = 1/(x+1)

du      -1
-- = ---------
dx   (x + 1)^2

Solve for dx:

dx = -(x + 1)^2 du
=============== END SUBSTITUTION BOX ================
Substitute into the integral.  You will find that the pieces that involve 'x' disappear nicely.  If they don't, you picked the wrong substitution, or the integral just can't be done this way.

(  e^(1/(x+1)           e^u (-(x + 1)^2) du      (
| ---------- dx = | ----------------------- = -1 | e^u

du
)  (x + 1)^2          (x + 1)^2                  )

That last matches a standard integral form.  At the end, substitute back for x's and you are done. I leave the rest to you.
----------------------


( ..sin^2 x-cos^2 x
..------------------ dx
)    ..sinx

I suggest that you do a little trig here.  Since cos^x = 1 - sin^x, the integral becomes:

(   2sin^2x - 1
| ------------- dx, and we divide:
)      sinx

(   
| (2 sin x - csc x) dx
)

The first term is easy, and I leave that to you.  To integrate EITHER of

(   
| sec x dx,  OR
)

(   
| csc x dx
)

there is a well-known trick:  Multiply top and bottom by

Sec x + tan x, for the first one, or
Csc x + cot x, for the second one, which we have here.

( csc^2 x + csc x cot x
| --------------------- dx
)   csc x + cot x

Amazingly, the top is (almost) exactly the derivative of the bottom.  (Almost, because there is a minus involved.  In the other one, it is exact -- no minus.)

So the derivative is  - ln| csc x + cot x|   

The minus comes from the fact that the derivatives of csc x and cot x have minus signs in them.
You can handle the rest.

--------------------------

(5e .. 1
   ..--- dx
)1 ..  x

This one is an easy integration;  1/x integrates to ln x.  (It's actually ln|x|, but your boundaries of integration are both positive.)

So the definite integral is

ln(5e) - ln(1) = ln 5 + ln e - ln 1 = ln 5 + 1 - 0 = ln

5 - 1
-------------------------

( .. x + 3
..-------- dx
) ..x^2 + 9

If you try to do this by substitution, it doesn't work out right away.  The derivative of the bottom, x, (Yes, I know that it is 2x, but that doesns't really matter.) is not a factor, but a term.  No good.

So what you have to do is separate the terms and integrate each as a problem in its own right.

First term:
(     x
| -------- dx
) x^2 + 9

This one can be handled by the substitution  u = x^2 + 9.  I leave that to you to finish up.

Second term:
(    3
| -------- dx
) x^2 + 9

When you see x^2 + a^2, in this case  x^2 + 3^2, think TRIGONOMETRIC SUBSTITION.  I suggest that you draw a right triangle, with theta as one angle.  When you see:

x^2 - a^2,  put x as the hypotenuse, a as the leg opposite theta.
a^2 - x^2,  put a as the hypotenuse, x as the leg opposite theta.
x^2 + a^2,  put x,a both as legs, with x opposite theta.

In this case, your triangle has  (writing t = theta)
x is the leg opposite t.
3 is the leg adjacent to t.
the hypotenuse is sqrt(x^2 + 9)

x = 3 tan t
dx = 3 sec^2 t

     3
------------- = cos t,  so x^2 + 9 = 9 sec^2 t
sqrt(x^2 + 9)

So now you can proceed to the substitution.


(    3
| -------- dx
) x^2 + 9

(    3
| ----------(3 sec^2 t) dt
) 9 sec^2 t

(  
|  dt = t = arctan(x/3)
)