calc 5.2b

Integrating trigonometric functions

Ex. 7 p.335 u-Substitution and the Log Rule

We can solve differential equations using the log rule as well.Solve the differential equation 1

ln

dy

dx x x

Solution - separate y things from x things and integrate both sides. Put the “plus C” on right side only.

1

lndy dx

x x 1

lndy dx

x x

1

lny dx

x x

There are three basic choices for u: u = x, u = x ln x, and u = ln x. The first two don’t fit the u’/u pattern. If I rewrite the function to be 1

lnxy dxx

the pattern fits because u = ln x and du = (1/x)dx

Rewrite with u-substitution:'u

y dxu

ln u C Back-substitute: ln lny x C

Up until now, we didn’t know how to integrate tan x, cot x, sec x, and csc x. With the Log rule, we can now do integration of these functions.

Ex 8 p. 336 Using a trig identity to integrate using log rule

Find tan x dxRewrite with trig identity:

sin

cos

xdx

x

Let u = cos x. Then du = – sin x dx

'udxu

ln cos x C

Ex 9 p. 336 Derivative of the secant functionThis problem needs a creative step, multiplying and dividing by the same quantity to make it work.

sec x dxFindsec tan

secsec tan

x xx dx

x x

2sec sec tan

sec tan

x x xdx

x x

Let u = sec x + tan x. (the denominator) Then du = sec x tan x + sec2 x, which is the numerator!

'udxu

Integrate and back-substitute

ln u C ln sec tanx x C

The other two are left as problems in the homework.

I remember these log ones by realizing if they are co- things, they have a negative in front. Secant and tangent go together, as do cosecant and cotangent.

These can be written in different forms, see #83-86

Ex 10 p. 337 Integrating Trigonometric Functions

Evaluate 22

4

1 cot x dx

Using Pythagorean Identity, 1 + cot2 x = csc2 x

22

4

csc x dx

2

4

csc x dx

2

4ln csc cotx x

ln 1 0 ln 2 1

ln csc cot ln csc cot2 2 4 4

ln 2 1 0.881372587

Be careful with parentheses if graphing and getting definite integral with the calculator.

ln 2 1

Last but not least, Ex 11 p. 337 Finding an average value

Recall that average value refers to average height, which would be area divided by width. 1

( )b

a

f x dxb a

Find the average value of f(x) = tan x on the interval [0, π/4]4

0

1tan

04

x dx

4ln cos ln cos 04

4 2ln ln 1

2

4 2ln

2

0.441

4

0

4ln cos x

1

π/4

y = 0.441 is avg value

5.2b p. 338 29-53 every other odd (skip 45), 65,70, 73, 83, 85