Question 355862
a) Hint: Draw a triangle with legs of {{{x}}}, {{{sqrt(1-x^2)}}} and with a hypotenuse of {{{1}}}. So {{{sin(theta)=x/1=x}}} ('x' is the leg opposite the angle {{{theta}}}). So because {{{x=sin(theta)}}}, *[Tex \LARGE dx=cos(\theta)d\theta]



So the integral then transforms to



*[Tex \LARGE \int \frac{\sin^2(\theta)}{\sqrt{1-\sin^2(\theta)}}\cos(\theta)d\theta]



But we know that *[Tex \LARGE \sqrt{1-\sin^2(\theta)}=\cos(\theta)], so...



*[Tex \LARGE \int \frac{\sin^2(\theta)}{\cos(\theta)}\cos(\theta)d\theta]



and then the cosine terms cancel to leave you with 



*[Tex \LARGE \int \sin^2(\theta)d\theta]



Let me know if this is enough to get you started.