If I = int e^{sin theta} (log sin theta + ope | vedclass

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(C) Let $\sin 	heta = t$. Then $\cos 	heta \, d	heta = dt$. Substituting these into the integral,we get: $I = \int e^t (\log t + \operatorname{cose

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$e^{\sin 	heta} (\log \sin 	heta - \operatorname{cosec} 	heta) + c$,(where $c$ is a constant of integration)

Vedclass · Answer

(C) Let $\sin 	heta = t$. Then $\cos 	heta \, d	heta = dt$. Substituting these into the integral,we get: $I = \int e^t (\log t + \operatorname{cosec}^2 	heta) \, dt$. Note that $\operatorname{cosec}^2 	heta = \frac{1}{\sin^2 	heta} = \frac{1}{t^2}$. So,$I = \int e^t (\log t + \frac{1}{t^2}) \, dt = \int e^t \log t \, dt + \int \frac{e^t}{t^2} \, dt$. Using integration by parts for $\int e^t \log t \, dt$: $\int e^t \log t \, dt = \log t \cdot e^t - \int \frac{e^t}{t} \, dt$. For $\int \frac{e^t}{t^2} \, dt$,using integration by parts (taking $u = \frac{1}{t^2}$ and $dv = e^t \, dt$): $\int \frac{e^t}{t^2} \, dt = \frac{e^t}{t^2} - \int e^t (-\frac{2}{t^3}) \, dt$ (This approach is complex). Alternatively,observe the form $\int e^t (f(t) + f'(t)) \, dt = e^t f(t) + c$. Let $f(t) = \log t - \frac{1}{t}$. Then $f'(t) = \frac{1}{t} + \frac{1}{t^2}$. This does not match directly. Let's re-evaluate: $I = \int e^t \log t \, dt + \int \frac{e^t}{t^2} \, dt = e^t \log t - \int \frac{e^t}{t} \, dt + \int \frac{e^t}{t^2} \, dt$. Actually,$\int e^t (\log t - \frac{1}{t}) \, dt = e^t \log t - \int \frac{e^t}{t} \, dt + \int \frac{e^t}{t} \, dt = e^t \log t$. Wait,the derivative of $e^t (\log t - \frac{1}{t})$ is $e^t (\log t - \frac{1}{t}) + e^t (\frac{1}{t} + \frac{1}{t^2}) = e^t (\log t + \frac{1}{t^2})$. Thus,$I = e^t (\log t - \frac{1}{t}) + c = e^{\sin 	heta} (\log \sin 	heta - \operatorname{cosec} 	heta) + c$.

If $I = \int e^{\sin \theta} (\log \sin \theta + \operatorname{cosec}^2 \theta) \cos \theta \, d\theta$,then $I$ is equal to

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