what is the integration of ∫cot⁡(ex)⋅exdxThis is my answer is it rightu=exdu=exdxdx=1exduThen∫cot⁡(u)⋅u1udu=∫cot⁡(u) du=csc2(u)

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what is the integration of ∫cot⁡(ex)⋅exdxThis is my answer is it rightu=exdu=exdxdx=1exduThen∫cot⁡(u)⋅u1udu=∫cot⁡(u) du=csc2(u)

PCCNQN4XKhjx · Accepted Answer

∫excot⁡{ex}dx=∫excos⁡{ex}sin⁡{ex}dx=∫1sin⁡{ex}d(sin⁡{ex})=ln|sin⁡{ex}|+C

Charlize Mora · Accepted Answer

No, your answer is not correct, but you were on the right track for the most part. Your problem was that you were looking at the wrong table when you were taking the integral of cot⁡u. Instead of the antiderivate of the cotangent function which is ∫cot⁡xdx=ln⁡{|sin⁡x|}+C , you used its derivative (ddxcot⁡x=−csc2x) to finish off the problem (you also dropped the minus sign for some reason). You got it completely backwards. Anyway, this is a simple integration by substitution (sometimes called u-substitution) problem. Integration by substitution is nothing more than the chain rule done backwards.∫cot⁡ex⋅exdx=u=ex⇒dudx=ddxex⇒dudx=ex⇒dx=duex=∫cot⁡u⋅uduu=∫cot⁡{u}du=ln⁡{|sin⁡{u}|}+C=ln⁡{|sin⁡{ex}|}+CAnswer: ln∣sin⁡ex∣+C

what is the integration of \(\displaystyle\int{\cot{{\left({e}^{{x}}\right)}}}\cdot{e}^{{x}}{\left.{d}{x}\right.}\) This is

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