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April 20, 2014

April 20, 2014

Posted by **Leanna** on Monday, April 11, 2011 at 10:25pm.

- Calculus -
**MathMate**, Monday, April 11, 2011 at 10:44pm1. ∫e^(lnx^2)dx

use the identity e^(ln(y)) = y to simplify the expression.

2. try the substitution u=sqrt(x).

- Calculus -
**Leanna**, Monday, April 11, 2011 at 11:00pmThanks for help on 1.

on 2. if i do u=sqrt(x) my du is 1/2x^(-1/2) and that means my du is in the denominator. So it would read 2integral of sin(u)/du

- Calculus -
**bobpursley**, Tuesday, April 12, 2011 at 12:27amNope, not on two.

You cant solve it that way easily.

This is difficult. Brake the sin function into its series equivalent, and integrate the series.

http://reference.wolfram.com/mathematica/ref/SinIntegral.html

- Calculus -
**bobpursley**, Tuesday, April 12, 2011 at 12:28amforget that last answer. I am tired.

- Calculus -
**MathMate**, Tuesday, April 12, 2011 at 5:14amFor 2, almost, but not quite!

Start with:

u=√x

du = (1/2)dx/√x

dx/√x = 2du

so

∫sin(√x) dx/√x

=∫sin(u)*2du

=-2cos(u)

=-2cos(√x)

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