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milkacha
integral of x^2*sqrt(1-x^2) dx using trig substitution
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.
I am up to sin^2 u cos^2 u once I rewrite cos^2 u in format 1-sin^2u I get sin^2u - sin^4 u du Now I am stuck.
see if below helps :- sin^2u cos^2u = 1/4 [ sin(2u) ]^2
Ok so you used a double-angle formula for sin2u. so now I have to simplift 1/4(sin2u)^2? Trig is my weakness - it's like Chinese :(
Ah yes, I am with you - the product formula!
Yes so to integrate that I get (1-cos4x)/2 dx using the sine half-angle formula. So I have the integral 1/4(1-cos4x / 2)dx I can replace the 1-cos4x with sin4x so then I have 1/4(sin4x/2)dx Where to from here? :/
And also I didn't see anywhere where I got rid of the dx = cos x du so there must be an error somewhere
\[\int\limits_{-}^{-}x ^{2}\sqrt{1-x ^{2}}dx\] let x=1 sin x Then: \[dx=\cos \theta d \theta; x ^{2}=\sin ^{2}\theta; 1-x ^{2}=\cos ^{2}\theta; \sqrt{\cos ^{2}\theta}=\cos \theta\] Then the original integral becomes \[\int\limits_{-}^{-}\sin ^{2}\theta*\cos \theta d \theta=\frac{ \sin ^{3}\theta }{ 3 }+C\]
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.)
Ok thanks I will that way.
I am still stuck on this one. The answer is: $\frac{x\sqrt{1-x^2}\left(2x^2-1\right)+\arcsin \left(x\right)}{8}+C$ I have NO idea how 1/4(sin^2 2theta) translates into this...... please help!!