Originally Posted by
Ackbeet So, number 4 works like this. You've got your integral,
$\displaystyle \int_{0}^{\sqrt{3}}\ln\left(\frac{2}{\sqrt{4-x^{2}}}\right)dx.$
Now then, suppose you draw your triangle with theta as the lower left angle. The squared terms have opposite signs, so you let 2 be the hypotenuse, and x the opposite side to theta. That means the adjacent side has length $\displaystyle \sqrt{4-x^{2}}$. You need to transform the limits, the integrand, and the differential so that there are no more x's, only theta's. Here are some relevant equations that let you do that:
$\displaystyle \frac{x}{2}=\sin(\theta)$
$\displaystyle \frac{2}{\sqrt{4-x^{2}}}=\sec(\theta)$.
So, hopefully, getting the integrand to have only theta's should be fairly straight-forward. How do you propose to get the limits and the differential to have only theta's?