3 dx V1 - 25x? 1. Cos ax dr = sin ax + C 2. sin ax dx -- cos ax + C %3D a a 4. csc ar de 1 = -- cot ax + C 3. ax dx = - tan ax + C a a 1 csc ax + C a 5. sec ax tan ax dx =- sec ax + C 6. csc ax cot ax= a dr = + C 1 + + C, b > 0, b + 1 In b 7. ar + C 8. h dr = a dx dr 9. at + x + C = sin- + C, a > 0 10. tan a %3! a* - x dx 11. -- - - seC + C, a > 0 xV a a a
3 dx V1 - 25x? 1. Cos ax dr = sin ax + C 2. sin ax dx -- cos ax + C %3D a a 4. csc ar de 1 = -- cot ax + C 3. ax dx = - tan ax + C a a 1 csc ax + C a 5. sec ax tan ax dx =- sec ax + C 6. csc ax cot ax= a dr = + C 1 + + C, b > 0, b + 1 In b 7. ar + C 8. h dr = a dx dr 9. at + x + C = sin- + C, a > 0 10. tan a %3! a* - x dx 11. -- - - seC + C, a > 0 xV a a a
3 dx V1 - 25x? 1. Cos ax dr = sin ax + C 2. sin ax dx -- cos ax + C %3D a a 4. csc ar de 1 = -- cot ax + C 3. ax dx = - tan ax + C a a 1 csc ax + C a 5. sec ax tan ax dx =- sec ax + C 6. csc ax cot ax= a dr = + C 1 + + C, b > 0, b + 1 In b 7. ar + C 8. h dr = a dx dr 9. at + x + C = sin- + C, a > 0 10. tan a %3! a* - x dx 11. -- - - seC + C, a > 0 xV a a a
Use a change of variables or Table to evaluate the following indefinite integrals. Check your work by differentiating.
Transcribed Image Text:3
dx
V1 - 25x?
Transcribed Image Text:1.
Cos ax dr =
sin ax + C
2. sin ax dx
-- cos ax + C
%3D
a
a
4. csc ar de
1
= -- cot ax + C
3.
ax dx = - tan ax + C
a
a
1
csc ax + C
a
5.
sec ax tan ax dx =- sec ax + C
6.
csc ax cot ax=
a
dr = + C
1
+ + C, b > 0, b + 1
In b
7.
ar + C
8.
h dr =
a
dx
dr
9.
at + x
+ C
= sin- + C, a > 0
10.
tan
a
%3!
a* - x
dx
11.
--
- - seC
+ C, a > 0
xV
a
a
a
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
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