Calculus
posted by Robert .
Hello, I just wanted to verify if my work was good.
Calculate the following integral by parts:
∫ upper limit is 1/5 and lower limit is 1/10. of 10sin^1 (5x)dx
so first I named the variables:
u = 10 sin^1 (5x)
du = 50 / sqr(125x^2)
dv = dx
v = x
so we get:
= 10 sin^1 (5x)(x)  ∫50x/(125x^2)
= 10 sin^1 (5x)(x)1/5, 1/10 
∫50x/(125x^2) 1/5, 1/10
let w = 125x^2
dw = 50xdx
= 10 sin^1 (5x)(x) + ∫ 1/sqr(w)dw
= 10 sin^1 (5x)(x) + 2sqr(w) + C 1/5, 1/10
= 180  (30 + 2sqr(0.75))
= 148.27
Thanks!

You're ok to this point:
10 sin^1 (5x)(x) + 2√(125x^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√(11)]  [10(1/10 * pi/6) + 2√(11/4)]
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/(125x^2)
I got
= 10 sin^1 (5x)(x)  ∫50x/(125x^2)^(1/2) dx
that last part can be integrated as
2(1  25x^2)^(1/2)
or 2√(125x^2)
so your final integral answer would be
10x sin^1 (5x)  2(125x^2)^(1/2)
see if that works for you. 
THanks Steve and Reiny!