Integral of the Day 8.5.24 | What Technique to Use? | Calculus 2 | Math with Professor V

Here's your latest Integral of the Day! It's a little spicier than the last two integrals, but still one that most of you should be able to work through and solve. Did you use a different approach? Do you like integrating inverse trig functions? Comment down below!

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[tonychow]

Hi Professor V, thank you for your new video. I let t^2=x and used parts and trig sub for the integral of vdu. I got the same result as yours. Cheers.

[aaravshah]

I did a u sub, then parts, then a trig sub!

[tonybluefor]

Let t=arcsin(sqr(x)) -->sin(t)=sqr(x) --> dx=2sin(t)cos(t)dt --> integral: 2tsin^2(t)cos(t)dt. By integration by part with u=t and dv=2sin^2(t)cos(t)dt and changing in the minus integral vdu sin^3(t) by (1-cos^2(t))sin(t), I got 2tsin^3(t)+2/3 cos(t) -2/9 cos^3(t) +C. With the right triangle we know that sin(t)=sqr(x), cos(t)=sqr(1-x) and t is already defined as arcsin(sqr(x)). It was a challenging integral😃