Homework Help: Calculus

Posted by Robert on Friday, September 7, 2012 at 9:44am.

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(1-25x^2)

dv = dx
v = x

so we get:

= 10 sin^-1 (5x)(x) - ∫50x/(1-25x^2)
= 10 sin^-1 (5x)(x)|1/5, 1/10 -
∫50x/(1-25x^2) |1/5, 1/10

let w = 1-25x^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!

Calculus - Steve, Friday, September 7, 2012 at 10:07am
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
Calculus - Reiny, Friday, September 7, 2012 at 10:15am
I think you dropped a square root and a dx in

so we get:

= 10 sin^-1 (5x)(x) - ∫50x/(1-25x^2)

I got
= 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)
or -2√(1-25x^2)

so your final integral answer would be

10x sin^-1 (5x) - 2(1-25x^2)^(1/2)

see if that works for you.
Calculus - Robert, Friday, September 7, 2012 at 7:57pm
THanks Steve and Reiny!

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