5.7 #53
We are wanting to compute
 tan(4  9) d .
First, we notice that there is no straightforward formula for computing the integral of tan( ) d . So, we are
going to have to algebraically manipulate the expression tan(4  9) d to get ourselves into a position where
we can straightforwardly compute the integral.
What can we algebraically do with the tangent function? A few identities come to mind: tan  
tan 2   1  sec 2  , and tan   1
cot 
sin 
,
cos 
. We don’t have a tangent squared, so that fact seems not to be the one
of choice. And, there is not a straightforward formula for integrating cotangent, so that fact seems to not be a
decent idea either. However, we do have formulas for integrating sine and cosine, let’s use that one.
sin(4  9)
sin(4  9)
, we get our integral to be in the form 
d . Now, we can
cos(4  9)
cos(4  9)
d
substitute. Thinking of d as
, we have a numerator expression of sin(4  9) d and a denominator
1
expression of cos(4  9) . Since we can get from sine to cosine or from cosine to sine via derivative, we want
to choose u to be equal to cos(4  9) so that the expression involving du involves the sine function.
Using tan(4  9) 
Letting u  cos(4  9) , we take derivative of both sides to get du  sin(4  9)  4 . Hence, our integral can be
written in the form
1 du
. We are now to a point where the integration is straight forward.
4 u
Just a remark. Another option is to immediately let u  4  9 and getting
1
tan(u) du , then using the fact
4
that tangent is sine over cosine. Either way, that tangent equals sine over cosine fact is the key in this case.
5.7 #59
We want to compute

sec 2 ( x )
x
dx . A key fact to recall here is that  sec 2 (u) du  tan(u)  C . So, if we can
work with those x and dx parts to simplify them down, we can get to a straightforward-to-integrate form.
Noting differences between th form we have and the form we want, most notably we can look inside the secant
function to note that we want x to be u .
Letting u  x
1
2
, we take derivative of both sides to get du 
into our integral expression, we get
1 12
1
x dx . Rewritten, this is
dx . Substituting
2
2 x
1
sec 2 (u) du , and now we can straightforwardly integrate.

2
5.1 #21 and #23
The derivative and the integral are indeed inverse operations. What one does, the other undoes. But, to see
more clearly what is going on in these problems, let’s consider this situation in general.
x
d  x
 10 f (t ) dt  . First, let’s use FTC: 10 f (t ) dt  F ( x)  F (10) .

dx 
Now, who cares exactly what f (t ) is and what the expression for F (x) is. The second term on the right-hand
side of the equal sign above is F (10) . This is just a number (ie. a constant). When you then apply the
d
d
derivative operation
to this expression, the  F (10) we be annihilated. What will remain is just
F (x) .
dx
dx
And, this of course simplifies down to just f (x) .
We are wanting to compute something like
For problems like 29-34 that are a bit trickier than this (where you’d be taking the derivative of expressions like
F ( x 2 ) and would have to use the chain rule when taking the derivative to get f ( x 2 )  2 x ), check out the
lecture commentary from Fri 1/29’s class. It provides more explanation on this matter.