Question

evaluate the integral cos(ln t)dt by parts?

Answer 1

Yeah, you'll want to approach this problem using integration by parts. However, this question is a little tricky and requires some clever manipulations:

First, set u=cos(ln(x)) and v'=1
that means that u' is given by:\[\frac{ -\sin( \ln x) }{ x }\] and v = x
So, using integration by parts we have:
\[\int\limits_{}^{}\cos(\ln x)dx = xcos(\ln x) - \int\limits_{}^{}x[\frac{ -\sin(\ln x) }{ x } ]dx\] which reduces to:
\[xcos(\ln(x)) +\int\limits_{}^{}\sin(\ln x)dx\] Now this might not not seem like much of an improvement just yet but wait--we haver to go through integration by parts a second time, giving us:
\[= xcos( \ln x) + x \sin (\ln x) - \int\limits_{}^{}\cos (\ln x)dx\] Now, if we add the integral of cos(ln x) to both sides we get:
\[2\int\limits_{}^{}\cos (\ln x)dx = x \cos (\ln x) + x \sin (\ln x)\] The last thing we do is divde by two, and you have your answer:
\[\int\limits_{}^{}\cos (\ln x)dx = \frac{ x }{ 2 }(\cos (\ln x)+\sin (\ln x))+C\]
Hope this helps!!