$\int \sin (\ln t) d t=\frac{t}{2}(\sin (\ln t)-\cos (\ln t))+C$ .

Name: $\int \sin (\ln t) d t=\frac{t}{2}(\sin (\ln t)-\cos (\ln t))+C$ .
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Question

Quizplus · Accepted Answer

This is a correct application of integration by parts, where one function is $u=\sin(\ln t)$ and the derivative of the other $dv=dt$, resulting in the given integral solution.

∫sin⁡(ln⁡t)dt=t2(sin⁡(ln⁡t)−cos⁡(ln⁡t))+C\int \sin (\ln t) d t=\frac{t}{2}(\sin (\ln t)-\cos (\ln t))+C∫sin(lnt)dt=2t​(sin(lnt)−cos(lnt))+C

$\int \sin (\ln t) d t=\frac{t}{2}(\sin (\ln t)-\cos (\ln t))+C$