# Integral Problems; I'm stumped.

• Dec 2nd 2008, 06:53 PM
Kari
Integral Problems; I'm stumped.
Integral (0, 2) of [(2)/(x+1) + (3)/(x-4)] dx

Integral (0, pi) of cos^2(x) sin^2(x) dx

• Dec 2nd 2008, 06:58 PM
Mathstud28
Quote:

Originally Posted by Kari
Integral (0, 2) of [(2)/(x+1) + (3)/(x-4)] dx

Integral (0, pi) of cos^2(x) sin^2(x) dx

For the first one try using this fact $\displaystyle \int\frac{u'}{u}du=\ln|u|+C$

For the second one consider the following

\displaystyle \begin{aligned}\cos^2(x)\sin^2(x)&=\left(\cos(x)\s in(x)\right)^2\\ &=\left(\frac{1}{2}\sin(2x)\right)^2\\ &=\frac{\sin^2(2x)}{4}\\ &=\frac{1-\cos(4x)}{8}\end{aligned}
• Dec 2nd 2008, 07:11 PM
Kari
Thanks!
• Dec 2nd 2008, 07:54 PM
Kari
Hello, I just finished the first problem I posted and was wondering if I could get it double checked? Thanks in advance!

Integral (0,2) of [(2/(x+1) + (3/(x-4)]dx

Integral (0,2) 2/(x+1) dx + Integral (0,2) 3/(x-4) dx

Let u = x+1; Let u = x-4;
du/dx = 1 du/dx = 1
du=dx du=dx

Integral (0,2) 2/u du + Integral (0,2) 3/u du

2*Integral 1/u du + 3*Integral 1/u du
2lnu + 3lnu
2ln(x+1) + 3ln(x-4)

Plugging in the intervals:

[2ln(2+1) + 3ln(2-4)] - [2ln(1) + 3ln(-4)]

[ln(3)^2 + ln(-2)^3] - [ln(1)^2 + ln(-4)^3]

[ln(9) + ln(-8)] - [ln(1) + ln(-64)]

ln(-72) - ln(-64)

ln(-72/-64)

ln (9/8) = 0.0511525224
• Dec 3rd 2008, 07:34 AM
Krizalid
Quote:

Originally Posted by Kari

[ln(9) + ln(-8)] - [ln(1) + ln(-64)]

ln(-72) - ln(-64)

ln(-72/-64)

ln (9/8) = 0.0511525224

May be you get the correct answer but your algebra is not correct, since, logarithms are defined for positive values, not for negative ones, hence, you should include the absolute value in each one of them.