Posted by Robert on Friday, September 7, 2012 at 9:44am.
You're ok to this point:
10 sin^-1 (5x)(x) + 2√(1-25x^2) + C |1/5, 1/10
By this time you should realize that radians are the measure of choice for trig stuff.
sin^-1(1/2) = pi/6
sin^-1(1) = pi/2
so you end up with
[10(1/5 * pi/2) + 2√(1-1)] - [10(1/10 * pi/6) + 2√(1-1/4)]
pi - (pi/6 + √3)
5pi/6 - √3
I think you dropped a square root and a dx in
so we get:
= 10 sin^-1 (5x)(x) - ∫50x/(1-25x^2)
= 10 sin^-1 (5x)(x) - ∫50x/(1-25x^2)^(1/2) dx
that last part can be integrated as
-2(1 - 25x^2)^(1/2)
so your final integral answer would be
10x sin^-1 (5x) - 2(1-25x^2)^(1/2)
see if that works for you.
THanks Steve and Reiny!
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