I'm sure how to go about solving this integral: ∫cos(ln(2x))dx Could someone explain how to set it up to use integration by parts?
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Use $\displaystyle u=\text{ln}(2x)$
Originally Posted by MaxJasper Use $\displaystyle u=\text{ln}(2x)$ Okay, that'll give me du = 1/x dx which I cannot substitute back into the original equation.
Originally Posted by Preston019 Okay, that'll give me du = 1/x dx which I cannot substitute back into the original equation. $\displaystyle u = \ln(2x)$ $\displaystyle x = \frac{e^u}{2}$ $\displaystyle dx = \frac{e^u}{2} \, du$ substitute ... $\displaystyle \int \cos(u) \cdot \frac{e^u}{2} \, du$ now perform the integration by parts ...
Originally Posted by skeeter $\displaystyle u = \ln(2x)$ $\displaystyle x = \frac{e^u}{2}$ $\displaystyle dx = \frac{e^u}{2} \, du$ substitute ... $\displaystyle \int \cos(u) \cdot \frac{e^u}{2} \, du$ now perform the integration by parts ... Oooooooohhh! Okay, that helped so much. Thanks!
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