$\displaystyle \int \frac{x^2}{\sqrt{4-x^2}} dx $ $\displaystyle \int sin^2(x)cos^3(x) dx $ I cant figure these 2 out. Can you also show the steps? Thank You so much
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I would use u substitution on the first one using: $\displaystyle u = 4-x^2$ and for the second one, try integration by parts: $\displaystyle \int udv = uv - \int vdu$
Originally Posted by larryboi7 $\displaystyle \int \frac{x^2}{\sqrt{4-x^2}} dx $ $\displaystyle \int sin^2(x)cos^3(x) dx $ I cant figure these 2 out. Can you also show the steps? Thank You so much use substitution method HINT : 1) let x= 2 sin t use sin^2(x)=1-cos^2(x) 2) $\displaystyle \int sin^2(x)cos^3(x) dx =\int sin^2(x) cos^2(x) cos(x)dx $ use cos^2(x)=1-sin^2(x) and let sin x=t so that cosx dx =dt and simplify
1) $\displaystyle x=2 \sin t \rightarrow dx=2 \cos t dt$
how do i find sin2t?
For the 2nd one, use cos(x)^2=1 - sin(x)^2 Then let u = sin x, and the remaining cos x is the du.
Originally Posted by danielomalmsteen 1) $\displaystyle x=2 \sin t \rightarrow dx=2 \cos t dt$ after doing all that, how do i solve for sin2t?
Originally Posted by larryboi7 after doing all that, how do i solve for sin2t? to restore the changes of variables?
Originally Posted by danielomalmsteen to restore the changes of variables? yes
$\displaystyle \sin {2t} = 2 \cdot \sin t \cdot \cos t = 2x \cdot \sqrt{1- \sin^2 t} = 2x \cdot \sqrt{1-\frac{x^2}{4}} = x \cdot \sqrt{4-x^2} $
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