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
$\displaystyle \begin{aligned}\cos^2(x)\sin^2(x)&=\left(\cos(x)\s in(x)\right)^2\\
&=\left(\frac{1}{2}\sin(2x)\right)^2\\
&=\frac{\sin^2(2x)}{4}\\
&=\frac{1-\cos(4x)}{8}\end{aligned}$
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