Beat Your Calculus Teacher for This Integral

In this video, I am evaluating this interesting integral using integration by parts and trig substitution.
#math #maths
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[@Min-cv7nt]

You are the best math professor on KZbin platform

[@iqtrainer]

Another cool video🎉

[@MrGLA-zs8xt]

Enjoyed this video so much professor

[@domedebali632]

I am pretty sure he is the most mathematically talented professor on KZbin

[@xinpingdonohoe3978]

Here's a cool trick. That +c can actually be useful. You set c=0, but try setting c=-1/2 to get (x²-1)/2 arcsin(x)+1/2 ∫(1-x²)/√(1-x²) dx =(x²-1)/2 arcsin(x)+1/2 ∫√(1-x²) dx Then just do an x=sin(u) substitution to get x²/2 arcsin(x)-1/4 arcsin(x)+1/4 x√(1-x²)+C It's the same method, but it uses that constant from integration by parts to speed things up.

[@NadiehFan]

This integral can be done using nothing but integration by parts. First, using integration by parts exactly as in the video we have (1) ∫ x∙arcsin(x)dx = ½x²∙arcsin(x) − ½∙∫ (x²/√(1 − x²))dx Now, since |x| ≤ 1 and therefore x²/√(1 − x²) = (1 − (1 − x²))/√(1 − x²) = 1/√(1 − x²) − (1 − x²)/√(1 − x²) = 1/√(1 − x²) − √(1 − x²) we have (2) ∫ (x²/√(1 − x²))dx = ∫ dx/√(1 − x²) − ∫ √(1 − x²)dx Again using integration by parts, we also have (3) ∫ √(1 − x²)dx = x√(1 − x²) + ∫ (x²/√(1 − x²))dx Substituting (3) in (2) we have ∫ (x²/√(1 − x²))dx = ∫ dx/√(1 − x²) − x√(1 − x²) − ∫ (x²/√(1 − x²))dx and therefore 2∙∫ (x²/√(1 − x²))dx = ∫ dx/√(1 − x²) − x√(1 − x²) which gives (4) ∫ (x²/√(1 − x²))dx = ½∙∫ dx/√(1 − x²) − ½∙x√(1 − x²) Substituting (4) in (1) we have ∫ x∙arcsin(x)dx = ½x²∙arcsin(x) − ½∙(½∙∫ dx/√(1 − x²) − ½∙x√(1 − x²)) which gives (5) ∫ x∙arcsin(x)dx = ½x²∙arcsin(x) + ¼∙x√(1 − x²) − ¼∙∫ dx/√(1 − x²) Finally, noting that 1/√(1 − x²)) is the derivative of arcsin(x) this gives (6) ∫ x∙arcsin(x)dx = ½x²∙arcsin(x) + ¼∙x√(1 − x²) − ¼∙arcsin(x) + C and we are done. This is equivalent with the result found in the video since we have ½∙sin(2∙arcsin(x)) = sin(arcsin(x))∙cos(arcsin(x)) = x√(1 − x²).

[@alexchan4226]

x^2*sin^（-1）x/2+c

[@Min-cv7nt]

Alex the crazy chan