Evaluate the integral: int sin(log x) dx | vedclass

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(A) Let $I = \int \sin(\log x) dx$. Substitute $\log x = t$, so $x = e^t$ and $dx = e^t dt$. Then $I = \int e^t \sin t dt$. Using the integration by p

Vedclass · Accepted Answer

$\frac{x}{2} [\sin(\log x) - \cos(\log x)] + c$

Vedclass · Answer

(A) Let $I = \int \sin(\log x) dx$. Substitute $\log x = t$, so $x = e^t$ and $dx = e^t dt$. Then $I = \int e^t \sin t dt$. Using the integration by parts formula $\int e^t \sin t dt = \frac{e^t}{1^2 + 1^2} (\sin t - \cos t) + c$. $I = \frac{e^t}{2} (\sin t - \cos t) + c$. Substituting back $t = \log x$ and $e^t = x$: $I = \frac{x}{2} [\sin(\log x) - \cos(\log x)] + c$.

Evaluate the integral: $\int \sin(\log x) dx$

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