Take the integral: integral x^2/sqrt(1-x^2) dx For the integrand x^2/sqrt(1-x^2), substitute x = sin(u) and dx = cos(u) du. Then sqrt(1-x^2) = sqrt(1-sin^2(u)) = cos(u) and u = sin^(-1)(x): = integral sin^2(u) du could someone explain how do you get integral sin^2(u) du in the end?
You require the double angular formula to solve that integral.
\[\int\limits_{{}}^{} \frac{ x^{2} }{ \sqrt{1-x^{2}} } dx \] wolfram says you substitute x = sin t and after differentiation you have dx = cos t dt
\[\Large \sin^2(x)=\frac{1}{2}(1-\cos(2x)) \]
ok i know how to solve integral of sin^2 (x) dx
but how did wolfram got there by using the substition method?
oh excuse me, thought that is your question.
well if \[\Large x=\sin\alpha \] Then you can differentiate that: \[\Large dx=\cos\alpha d\alpha \] and now plug everything back into your original integral.
\[\Large \int \frac{\sin^2\alpha}{\cos\alpha}\cdot \cos\alpha d\alpha=\int\sin^2\alpha d\alpha \]
oh yeah... thanks for clarifying . i got silly on the way of finding the final result cause the initial integral is something like integral from -1/2 to 1/2 of x^2*sqrt (1-x^2) dx
thanks again.
you're very welcome
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