If int e^{sin x}(1+sec x tan x) d x=e^{sin x} | vedclass

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(C) Given the integral $\int e^{\sin x}(1+\sec x 	an x) d x = e^{\sin x} f(x) + c$. Let $I = \int e^{\sin x}(1+\sec x 	an x) d x$. We can rewrite th

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$2$

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(C) Given the integral $\int e^{\sin x}(1+\sec x 	an x) d x = e^{\sin x} f(x) + c$. Let $I = \int e^{\sin x}(1+\sec x 	an x) d x$. We can rewrite the integrand as $e^{\sin x} + e^{\sin x} \sec x 	an x$. This does not immediately fit the form $\int e^{g(x)}(g'(x) + h(x)) dx$. Let us differentiate $e^{\sin x} f(x)$ with respect to $x$: $\frac{d}{dx} [e^{\sin x} f(x)] = e^{\sin x} \cos x f(x) + e^{\sin x} f'(x) = e^{\sin x} (f'(x) + f(x) \cos x)$. Comparing this with the integrand $e^{\sin x}(1+\sec x 	an x)$,we have $f'(x) + f(x) \cos x = 1 + \sec x 	an x$. By inspection,if $f(x) = \sec x$,then $f'(x) = \sec x 	an x$. Substituting this into the equation: $\sec x 	an x + \sec x \cos x = \sec x 	an x + 1$. This matches the integrand. Thus,$f(x) = \sec x$. We need to find the number of solutions for $f(x) = 1$ in $0 \leq x \leq 2 \pi$. $\sec x = 1 \implies \cos x = 1$. In the interval $[0, 2 \pi]$,$\cos x = 1$ at $x = 0$ and $x = 2 \pi$. Thus,there are $2$ solutions.

If $\int e^{\sin x}(1+\sec x \tan x) d x=e^{\sin x} f(x)+c$,then in $0 \leq x \leq 2 \pi$,the number of solutions of $f(x)=1$ is

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