TheBabe Group Title Little Review: what is the integral of sin^3(2x) 3 years ago 3 years ago

1. TheBabe

all right, $\sin^3(2x)$

2. TheBabe

seems like i need to do substitution rule, then use trig identities?

3. LeoMessi

right, let $u=2x$, so $du = 2 dx$, so $dx=(1/2)du$

4. TheBabe

$\int\limits_{}^{}\sin^3 x$

5. LeoMessi

forgot the dx ;), so that is equal to $(1/2)\int\limits_{}^{}\sin^3(u)du$

6. LeoMessi

$sin^2x=1-cos^2x$, so we'll have $(1/2)\int\limits_{}^{}\sin^3(u)du$ = $(1/2)\int\limits_{}^{}\sin^2(u)*sin(u)du$ = $(1/2)\int\limits_{}^{}(1-sin^2(u))sin(u)du$

7. LeoMessi

so we can apply substitution rule again for $v=\sin(u)$

8. LeoMessi

$dv = \cos(u)du$, so $(1/2)\int\limits_{}^{}(1-v^2)dv$

9. TheBabe

wait, shouldn't it be 1-cos^2(u) above?

10. LeoMessi

ahhh, yesss... sorry: $(1/2)\int\limits_{}^{}(1-cos^2(u))sin(u)du$

11. TheBabe

so v=cos(u), which is dv=-sin(u)du

12. TheBabe

i think I got it now: so$(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du = (-1)*(1/2) \int\limits_{}^{} (1-v^2)dv = (-1)(1/2) [ v-(1/3)v^3 + C]$

13. TheBabe

i think I got it now: so$(1/2)\int\limits\limits_{}^{}(1-\cos^2(u))\sin(u)du$ = $(-1)*(1/2) \int\limits_{}^{} (1-v^2)dv$ = $(-1)(1/2) [ v-(1/3)v^3 + C]$

14. TheBabe

so plug back in and get v = cos u --> cos (2x), $(-1)(1/2) [ v-(1/3)v^3 + C] = (-1)(1/2) [ \cos(2x) - (1/3)\cos^3(2x) + C]$ $= (-1)(1/2) [ cos(2x) - (1/3)cos^3(2x) + C]$