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int_{0}^{3} x/(sqrt{x^2+16}
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let\[x =\tan \theta\]
:) thx
On closer inspection, you don't need trig sub, u sub would suffice. I never noticed until after I worked it out myself. But hey, good practice for trig sub. When you get it on the test you going to knock it out of the park.
shouldnt x = 4tan(t)? that way; when you square it you get: sqrt(16 tan^2(t) + 16) = sqrt(16(tan^2+1))) = 4sec^2
err.... 4 sec :)
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Yes. I think he picked that up.
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