I have my Final Exam tomorrow and would like to know if these are the correct answers to my review problems. The teacher did not list which problems to solve using trigonometric substitution or regular substitution. How do I know which problems to use the trig sub. Even if you can't provide an answer for me, I would still like to know if I have each one right or wrong. Thankyou in advance. I couldn't figure out how to make a integral symbol for each problem, sorry.
1. sin (2x+1) dx
s=2x+1
ds=2dx
= 1/2 sin (s) ds
= sin (s) is -cos (s)
= -cos (s)/2 + C
= -1/2 cos (2x+1) + C
2. (2x+5)^5 dx
s=2x+5
ds= 2dx
= 1/2 s^5 ds
= s^6/12 + C
= 1/12 (2x+5)^6 + C
3. sqrt [2x-1] dx
s= 2x-1
ds= 2dx
= 1/2 sqrt ds
= s^(3/2) / 3 + C
= 1/3 (2x-1)^3/2 + C
4. x sqrt [1+x^2] dx
s=x^2+1
ds=2xdx
= 1/2 sqrt ds
= s^3/2 / 3 + C
= 1/3 (x^2+1)^3/2 + C
5. r^2/sqrt [1-r^3] dr
s=1-r^3
ds=-3r^2 dr
=-2/3 sqrt [1-r^3] + C
6. sin[x] cos[x] dx
s=cos (x)
ds=-sin(x) dx
= - (s) ds
= -s^2/2 + C
= 1/2 cos(x)^2 + C
7. sin^3x cos(x) dx
s=sinx
ds=cos(x)dx
= 1/3 s^3 ds
= sin (x)^4/4 +C
1. sin (2x+1) dx
s=2x+1
ds=2dx
= 1/2 sin (s) ds
= sin (s) is -cos (s)
= -cos (s)/2 + C
= -1/2 cos (2x+1) + C
2. (2x+5)^5 dx
s=2x+5
ds= 2dx
= 1/2 s^5 ds
= s^6/12 + C
= 1/12 (2x+5)^6 + C
3. sqrt [2x-1] dx
s= 2x-1
ds= 2dx
= 1/2 sqrt
= s^(3/2) / 3 + C
= 1/3 (2x-1)^3/2 + C
4. x sqrt [1+x^2] dx
s=x^2+1
ds=2xdx
= 1/2 sqrt
= s^3/2 / 3 + C
= 1/3 (x^2+1)^3/2 + C
5. r^2/sqrt [1-r^3] dr
s=1-r^3
ds=-3r^2 dr
=-2/3 sqrt [1-r^3] + C
6. sin[x] cos[x] dx
s=cos (x)
ds=-sin(x) dx
= - (s) ds
= -s^2/2 + C
= 1/2 cos(x)^2 + C
7. sin^3x cos(x) dx
s=sinx
ds=cos(x)dx
= 1/3 s^3 ds
= sin (x)^4/4 +C