# 1 Quick integration problem

• Dec 1st 2009, 05:51 PM
Lord Darkin
1 Quick integration problem
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)
• Dec 1st 2009, 06:07 PM
skeeter
Quote:

Originally Posted by Lord Darkin
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 what is this ?

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)

$\int_1^4 \frac{1}{\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx$

$u = \frac{\pi \sqrt{x}}{2}$

$du = \frac{\pi}{4\sqrt{x}} \, dx$

$\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 ...

$\frac{4}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos{u} \, du$

finish.
• Dec 1st 2009, 06:45 PM
Lord Darkin
Quote:

Originally Posted by skeeter
$\int_1^4 \frac{1}{\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx$

$u = \frac{\pi \sqrt{x}}{2}$

$du = \frac{\pi}{4\sqrt{x}} \, dx$

$\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 ...

$\frac{4}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos{u} \, du$

finish.

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
• Dec 2nd 2009, 05:33 AM
skeeter
Quote:

Originally Posted by Lord Darkin
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 $u$ in this case.

the antiderivative of $\cos{u}$ is $\sin{u}$ , evaluated from $\frac{\pi}{2}$ to $\pi$ using the Fundamental Theorem of Calculus.

$\sin(\pi) - \sin\left(\frac{\pi}{2}\right) = 0 - 1 = -1$

therefore, the value of the original definite integral is $-\frac{4}{\pi}$

you need to review/relearn the process of finding the values of definite integrals by substitution.