If f(x) = cos x, then the value of int frac{e | vedclass

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(B) Given $f(x) = \cos x$, then $f'(x) = -\sin x$. The integral is $I = \int \frac{e^{\cos x}(x \sin^3 x + \cos x)}{1 - \cos^2 x} dx$.Since $1 - \cos^

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$e^{\cos x}(	ext{cosec } x + x) + c$

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(B) Given $f(x) = \cos x$, then $f'(x) = -\sin x$. The integral is $I = \int \frac{e^{\cos x}(x \sin^3 x + \cos x)}{1 - \cos^2 x} dx$.Since $1 - \cos^2 x = \sin^2 x$, we have $I = \int e^{\cos x} \left( \frac{x \sin^3 x}{\sin^2 x} + \frac{\cos x}{\sin^2 x} ight) dx$.$I = \int e^{\cos x} (x \sin x + \cot x 	ext{cosec } x) dx$.Let $u = e^{\cos x}$, then $du = -\sin x e^{\cos x} dx$, so $e^{\cos x} dx = -\frac{du}{\sin x}$. This approach is complex. Instead, use integration by parts.Let $I = \int e^{\cos x} x \sin x dx + \int e^{\cos x} \cot x 	ext{cosec } x dx$.For the first part, let $u = x$ and $dv = e^{\cos x} \sin x dx$. Then $du = dx$ and $v = -e^{\cos x}$.$\int x e^{\cos x} \sin x dx = -x e^{\cos x} - \int (-e^{\cos x}) dx = -x e^{\cos x} + \int e^{\cos x} dx$.This does not simplify easily. Re-evaluating: $I = \int e^{\cos x} (x \sin x + 	ext{cosec } x \cot x) dx$.Note that $\frac{d}{dx}(e^{\cos x} \cdot x) = e^{\cos x} - x \sin x e^{\cos x}$ and $\frac{d}{dx}(e^{\cos x} 	ext{cosec } x) = e^{\cos x} (-\sin x 	ext{cosec } x - 	ext{cosec } x \cot x) = e^{\cos x} (-1 - 	ext{cosec } x \cot x)$.Actually, $\frac{d}{dx}(e^{\cos x}(	ext{cosec } x + x)) = e^{\cos x}(-\sin x 	ext{cosec } x - 	ext{cosec } x \cot x + 1 + x(-\sin x)) = e^{\cos x}(-1 - 	ext{cosec } x \cot x + 1 - x \sin x) = -e^{\cos x}(	ext{cosec } x \cot x + x \sin x)$.Thus, $\int e^{\cos x}(x \sin x + 	ext{cosec } x \cot x) dx = -e^{\cos x}(	ext{cosec } x + x) + c$.

If $f(x) = \cos x$, then the value of $\int \frac{e^{f(x)}(x \sin^3 x + f(x))}{1 - (f(x))^2} dx =$

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