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integation {e^(log x)+sin x}cos x dx
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try multiplying through by cos(x) and then integrating each of the components separately.
This is your integral?\[\large\rm \int\limits (e^{\ln x}+\sin x)\cos x~dx\]
yes it is.
@zepdrix
Recall that the exponential and log are inverse operations of one another, so taking their composition will give us the argument back as a result,\[\large\rm e^{\ln x}=x\]
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\[\large\rm \int\limits\limits (x+\sin x)\cos x~dx\]
Distribute the cosx as Luff indicated,\[\large\rm \int\limits x \cos x~dx+\int\limits \sin x \cos x~dx\]
ahh! got it!! thank you @zepdrix
Got it? :) Cool
thanks!
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\(\Large\rm \color{royalblue}{\text{Welcome to OpenStudy! :)}}\)
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