int frac{1+sin (log x)}{1+cos (log x)} d x= | vedclass

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(D) Let $I = \int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int \fra

Vedclass · Accepted Answer

$x \cdot 	an \left(\frac{\log x}{2}ight)+c$,where $c$ is a constant of integration.

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(D) Let $I = \int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int \frac{1+\sin t}{1+\cos t} e^t dt$. Using trigonometric identities $1+\cos t = 2\cos^2(t/2)$ and $\sin t = 2\sin(t/2)\cos(t/2)$,we get: $I = \int \frac{\sin^2(t/2) + \cos^2(t/2) + 2\sin(t/2)\cos(t/2)}{2\cos^2(t/2)} e^t dt$. $I = \frac{1}{2} \int (	an^2(t/2) + 1 + 2	an(t/2)) e^t dt$. $I = \frac{1}{2} \int (\sec^2(t/2) + 2	an(t/2)) e^t dt$. Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = 	an(t/2)$ and $f'(t) = \frac{1}{2}\sec^2(t/2)$: $I = e^t 	an(t/2) + c$. Substituting $t = \log x$ back,we get $I = x 	an \left(\frac{\log x}{2}ight) + c$.

$\int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x=$

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