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Find the integral: $\int \sin(\log x) \, dx$ solution

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Csc, sec, and cot of special angles

Full solution

Q. Find the integral:

\int \sin(\log x) \, dx

solution

Use substitution: Use substitution. Let $u = \log(x)$ , then $du = \frac{1}{x} dx$ or $dx = x du$ .
Substitute and integrate: Since $x = e^u$ , substitute $x$ and $dx$ in the integral. $\newline$ $\int \sin(\log(x)) dx = \int \sin(u) e^u du$
Integration by parts: Use integration by parts. Let $v = \sin(u)$ and $dw = e^u du$ . Then $dv = \cos(u) du$ and $w = e^u$ .
Apply integration by parts: Apply the integration by parts formula $\int v dw = vw - \int w dv$ . $\newline$ $\int \sin(u) e^u du = \sin(u) e^u - \int e^u \cos(u) du$
Use integration by parts again: Use integration by parts again on $\int e^u \cos(u) du$ . Let $v = \cos(u)$ and $dw = e^u du$ . Then $dv = -\sin(u) du$ and $w = e^u$ .
Combine results: Apply the integration by parts formula again. $\newline$ $\int e^u \cos(u) du = \cos(u) e^u - \int e^u (-\sin(u)) du$ $\newline$ $= \cos(u) e^u + \int e^u \sin(u) du$
Solve for integral: Combine the results from Steps $4$ and $6$ . $\newline$ $\int \sin(u) e^u du = \sin(u) e^u - (\cos(u) e^u + \int e^u \sin(u) du)$ $\newline$ $= \sin(u) e^u - \cos(u) e^u - \int e^u \sin(u) du$
Substitute back: Solve for $\int e^u \sin(u) du$ . $\newline$ $\int e^u \sin(u) du + \int e^u \sin(u) du = \sin(u) e^u - \cos(u) e^u$ $\newline$ $2 \int e^u \sin(u) du = \sin(u) e^u - \cos(u) e^u$ $\newline$ $\int e^u \sin(u) du = \frac{1}{2} (\sin(u) e^u - \cos(u) e^u)$
Substitute back: Solve for $\int e^u \sin(u) du$ . $\newline$ $\int e^u \sin(u) du + \int e^u \sin(u) du = \sin(u) e^u - \cos(u) e^u$ $\newline$ $2 \int e^u \sin(u) du = \sin(u) e^u - \cos(u) e^u$ $\newline$ $\int e^u \sin(u) du = \frac{1}{2} (\sin(u) e^u - \cos(u) e^u)$ Substitute back $u = \log(x)$ . $\newline$ $\int \sin(\log(x)) dx = \frac{1}{2} (\sin(\log(x)) x - \cos(\log(x)) x) + C$

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