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

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

Printable View

- Dec 2nd 2008, 06:53 PMKariIntegral 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 PMMathstud28
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 PMKari
Thanks!

- Dec 2nd 2008, 07:54 PMKari
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 AMKrizalid