int frac{1}{log a} (a^x cos a^x) , dx = | vedclass

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Question

(C) Let $I = \int \frac{1}{\log a} (a^x \cos a^x) \, dx$. Substitute $t = a^x$. Then,differentiating with respect to $x$,we get $dt = a^x \log a \, dx

Vedclass · Accepted Answer

$\frac{1}{(\log a)^2} \sin a^x + c$

Vedclass · Answer

(C) Let $I = \int \frac{1}{\log a} (a^x \cos a^x) \, dx$. Substitute $t = a^x$. Then,differentiating with respect to $x$,we get $dt = a^x \log a \, dx$,which implies $a^x \, dx = \frac{dt}{\log a}$. Substituting these into the integral: $I = \int \frac{1}{\log a} \cos t \left( \frac{dt}{\log a} \right) = \frac{1}{(\log a)^2} \int \cos t \, dt$. Integrating $\cos t$ gives $\sin t$: $I = \frac{1}{(\log a)^2} \sin t + c$. Substituting back $t = a^x$,we get: $I = \frac{1}{(\log a)^2} \sin a^x + c$.

$\int \frac{1}{\log a} (a^x \cos a^x) \, dx = $

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