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## TheBabe Little Review: what is the integral of sin^3(2x) 2 years ago 2 years ago Edit Question Delete Cancel Submit

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

• 2 years ago
2. TheBabe
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seems like i need to do substitution rule, then use trig identities?

• 2 years ago
3. LeoMessi
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right, let $u=2x$, so $du = 2 dx$, so $dx=(1/2)du$

• 2 years ago
4. TheBabe
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$\int\limits_{}^{}\sin^3 x$

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

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

• 2 years ago
7. LeoMessi
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so we can apply substitution rule again for $v=\sin(u)$

• 2 years ago
8. LeoMessi
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$dv = \cos(u)du$, so $(1/2)\int\limits_{}^{}(1-v^2)dv$

• 2 years ago
9. TheBabe
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wait, shouldn't it be 1-cos^2(u) above?

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

• 2 years ago
11. TheBabe
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so v=cos(u), which is dv=-sin(u)du

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

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

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

• 2 years ago
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