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hartnn
 one year ago
Best ResponseYou've already chosen the best response.2then 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.

milkacha
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

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

milkacha
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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 :(

milkacha
 one year ago
Best ResponseYou've already chosen the best response.0Ah yes, I am with you  the product formula!

milkacha
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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? :/

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

mathmale
 one year ago
Best 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 \[\int\limits_{}^{}\sin ^{2}\theta*\cos \theta d \theta=\frac{ \sin ^{3}\theta }{ 3 }+C\]

mathmale
 one year ago
Best ResponseYou've already chosen the best response.0Since 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.)

milkacha
 one year ago
Best ResponseYou've already chosen the best response.0Ok thanks I will that way.

milkacha
 one year ago
Best ResponseYou've already chosen the best response.0I 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!!
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