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milkacha
Group Title
integral of x^2*sqrt(1x^2) dx
using trig substitution
 10 months ago
 10 months ago
milkacha Group Title
integral of x^2*sqrt(1x^2) dx using trig substitution
 10 months ago
 10 months ago

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hartnn Group TitleBest ResponseYou've already chosen the best response.2
did u try x = sin u ??
 10 months ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
then you can always write sin^2 u cos^2 u du as 1/4 (sin^2 2u)du and integrating square of sin function is a pretty standard procedure. let me know if you get stuck in any step.
 10 months ago

milkacha Group TitleBest ResponseYou've already chosen the best response.0
I am up to sin^2 u cos^2 u once I rewrite cos^2 u in format 1sin^2u I get sin^2u  sin^4 u du Now I am stuck.
 10 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
see if below helps : sin^2u cos^2u = 1/4 [ sin(2u) ]^2
 10 months ago

milkacha Group TitleBest ResponseYou've already chosen the best response.0
Ok so you used a doubleangle formula for sin2u. so now I have to simplift 1/4(sin2u)^2? Trig is my weakness  it's like Chinese :(
 10 months ago

milkacha Group TitleBest ResponseYou've already chosen the best response.0
Ah yes, I am with you  the product formula!
 10 months ago

milkacha Group TitleBest ResponseYou've already chosen the best response.0
Yes so to integrate that I get (1cos4x)/2 dx using the sine halfangle formula. So I have the integral 1/4(1cos4x / 2)dx I can replace the 1cos4x with sin4x so then I have 1/4(sin4x/2)dx Where to from here? :/
 10 months ago

milkacha Group TitleBest ResponseYou've already chosen the best response.0
And also I didn't see anywhere where I got rid of the dx = cos x du so there must be an error somewhere
 10 months ago

mathmale Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}x ^{2}\sqrt{1x ^{2}}dx\] let x=1 sin x Then: \[dx=\cos \theta d \theta; x ^{2}=\sin ^{2}\theta; 1x ^{2}=\cos ^{2}\theta; \sqrt{\cos ^{2}\theta}=\cos \theta\] Then the original integral becomes \[\int\limits_{}^{}\sin ^{2}\theta*\cos \theta d \theta=\frac{ \sin ^{3}\theta }{ 3 }+C\]
 10 months ago

mathmale Group TitleBest ResponseYou've already chosen the best response.0
Since x=sin theta, x/1=sin theta, and theta = \[\theta=\sin ^{1}\frac{ x }{ 1 }.\] Substitute this into the prior expression to obtain the integral in terms of x. (Result is mighty simple.)
 10 months ago

milkacha Group TitleBest ResponseYou've already chosen the best response.0
Ok thanks I will that way.
 10 months ago

milkacha Group TitleBest ResponseYou've already chosen the best response.0
I am still stuck on this one. The answer is: $\frac{x\sqrt{1x^2}\left(2x^21\right)+\arcsin \left(x\right)}{8}+C$ I have NO idea how 1/4(sin^2 2theta) translates into this...... please help!!
 10 months ago
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