Solution to Exercise 11.5-2 © 2001 by Karl Hahn

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The problem was to use trig substitution to find the indefinite integral:

____________ √35 + 2x - x2 dx

Step 1: Complete the square. This is a negative version, so the quadratic is

-q(x) = 35 + 2x - x2

or

q(x) = x2 - 2x - 35

Adding half the middle term to x to get the first substitution, v, gives

v = x - 1

and

v2 = x2 - 2x + 1

Now what do you have to add to v² to get it to equal the original quadratic,  q(x)? 

    v2           =    x2  -  2x  +   1
        -  36    =               -  36
                                      
    v2  -  36    =    x2  -  2x  -  35    =    q(x)

Since you've got a negative version, you will substitute with

   -q(x)  =  35  +  2x  -  x2  =  36  -  v2

Notice also that since  v = x - 1,  then  dv = dx. 

____________ √35 + 2x - x2 dx

=

_______ √36 - v2 dv

And from here you integrate the v integral exactly as you integrated the x integral in exercise 1.

Step 2: Factor out the 36. Pull it all the way out of the radical, hence it will have to get square-rooted out there

6

1 -

v2 36

dv

or equivalently

6

1 -

v 6

2

dv

Step 3: Make the substitution. For this example the substitution (and taking its derivative implicitly) is:

v sin(u) = 6

and

du 1 cos(u) = dv 6

or

6 cos(u) du = dv

When you put that in for v and dv, the integral becomes

_______ √36 - v2 dv

= 36

___________ √1 - sin2(u) cos(u) du =

36

cos2(u) du

Step 4: Integrate. We already did the integral of cosine squared in the main text, so click here if you don't remember the trick. You will find that

36

cos2(u) du = 36

1 1 u + sin(u)cos(u) 2 2

+ C

Step 5: Substitute back to v. You have to take the inverse function on the v-to-u substitution to substitute back the bare u, so it becomes  arcsin(v/6).  The  sin(u) becomes  v/6.  And the  cos(u)  is the same as

    ___________
   √1 - sin2(u)

so it becomes

v2 1 − 36

Putting all the pieces together and multiplying the 36 through both the terms, you get

_______ √36 - v2 dv

36 = arcsin 2

v 6

6 + v 2

v2 1 − 36

+ C =

36 arcsin 2

v 6

1 + v 2

_______ √36 - v2

+ C

Step 6: Substitute back to x. Your first substitution was  v = x - 1.  So put that into the above:

____________ √35 + 2x - x2 dx

=

36 arcsin 2

x - 1 6

6 + (x - 1) 2

(x - 1)2 1 − 36

+ C =

36 arcsin 2

x - 1 6

1 + (x - 1) 2

____________ √35 + 2x - x2

+ C

Now very carefully (paying close attention to what all the constant factors do) take the derivative above and confirm that you get back the original integrand.

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