<p>Evaluate the integral of the following (integration boundaries are from x=2 to x=6): </p>
<p>(6+x)^(1/2)
------------ dx
(6-x)^(1/2)</p>
<p>answer: 3pie-6arcsin(1/3)+(32)^(1/2)</p>
<p>No matter how much I try, I can't solve it. :(
I really hope questions like that won't show up on the actual exam.</p>
<p>Thanks.</p>
<p>from a cursory glance, i would use the antiderivs of inverse trig functions.</p>
<p>multiply by (6+x)^(1/2) to the numerator and the denominator, and then split up the integral. Use u-sub on one part and arcsin integral for the second. </p>
<p>Hopefully that helped even though its hard to describe it in words</p>
<p>A good rule to remember is: If you can't find a way to integrate it, multiply by the conjugate.</p>
<p>So, multiply top and bottom by either √(6-x) or √(6+x). Then it becomes obvious you need to use trigonometric substitution w/ x=6sinθ. Use identity 1-sin^2θ=cos^2θ and solve. So... it goes to ∫(secθ+tanθ)dx, you do the antiderivative, un-substitute, and solve.</p>