Page 226 - Calculus Workbook For Dummies

210 Part IV: Integration and Infinite Series

√9 + 16x 2 4x

θ

3
4 x
i
3. Solve tan = for x, differentiate, and solve for dx.
3
4 x = 3 tani dx = 3 sec i
2
di 4
3
x = tani 3 sec i di
2
4 dx = 4
4. Do the a thing.
9 + 16 x 2 seci 2 seci
3 = 9 + 16 x = 3
5. Substitute. Hint: There are three substitutions here, not just two like in the example.
# x 9 + 16 x dx
2
3 3 2
= # c tani ^ m 3 seci c h sec i dim
4 4
27 # 3
i
= tan sec i di
16
6. Now you’re back in trigonometric integral territory. Split off a sec tani factor.
i
27 # 2
i
= sec i ^ sec tanih di
16
7. Integrate. This is in # u du form, so = 27 1 sec im + C = 9 sec i + C .
3
2
3
c
16 3 16
J 2 N 3
9 K 9 + 16 x O 1 2 / 3 2
8. Switch back to x: = 16 K 3 O + C = 48 _ 9 + 16 x i + C .
L P
n # dx = x + C .
2
2
2
_ 9 x + 4i 9 x + 4 4 9 x + 4
1. Rewrite as # dx 3 .
2
3 ^ x + 2 2
h
u
2. Draw your triangle, remembering that tan =i a . See the following figure.
√9x 2 + 4 3x

θ

2
3 x
i
3. Solve tan = for x, differentiate, and solve for dx.
2
2 2 2
3 x = 2 tani x = tani dx = sec i di
3 3