# Thread: Integral Problems; I'm stumped.

1. ## 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

2. 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}

3. Thanks!

4. 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

5. 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.