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Integral of (1/sqrt(x)) * cos((pi*sqrt(x))/2) in the range of 1 to 4. I'm supposed to integrate by substitution using u = pi*sqrt(x) all divided by 2
Thus,
dy = (pi*x^3/2)/3
And I get
integral of:
3/pi * cos(u) du from the range of 1 to 4, but this does not take care of the problem with the 1/sqrt(x) in the original and the new x^(3/2) in the new one. (Crying)
$\displaystyle \int_1^4 \frac{1}{\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx$Quote:
Originally Posted by Lord Darkin
$\displaystyle u = \frac{\pi \sqrt{x}}{2}$
$\displaystyle du = \frac{\pi}{4\sqrt{x}} \, dx$
$\displaystyle \frac{4}{\pi} \int_1^4 \frac{\pi}{4\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx
$
substitute and reset the limits of integration ...
$\displaystyle \frac{4}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos{u} \, du$
finish.
Quote:
Originally Posted by skeeter
Ohhh my du was wrong. You got the right answer according to my answer key. Thanks. I forgot that I had to take the derivative of u, not take the integrand.
Last question - would this still work if I resetted th limits to be 0 and (-1/2)? Cos(pi) = -1/2 and cos(pi/2) = 0
no, the limits of integration are related to the variable of integration which is $\displaystyle u$ in this case.Quote:
Originally Posted by Lord Darkin
the antiderivative of $\displaystyle \cos{u}$ is $\displaystyle \sin{u}$ , evaluated from $\displaystyle \frac{\pi}{2}$ to $\displaystyle \pi $ using the Fundamental Theorem of Calculus.
$\displaystyle \sin(\pi) - \sin\left(\frac{\pi}{2}\right) = 0 - 1 = -1$
therefore, the value of the original definite integral is $\displaystyle -\frac{4}{\pi}$
you need to review/relearn the process of finding the values of definite integrals by substitution.
