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Question: If $F(x)=\int \frac{e^{\cos x}(1+x\sin x)}{x^2}dx$ and $F(\pi)=\frac{-1}{e\pi}$, then $F(\frac{\pi}{...

If F(x)=∫ecos⁡x(1+xsin⁡x)x2dxF(x)=\int \frac{e^{\cos x}(1+x\sin x)}{x^2}dx and F(π)=−1eπF(\pi)=\frac{-1}{e\pi}, then F(π3)=F(\frac{\pi}{3})=

A

π3e\frac{\pi}{3\sqrt{e}}

B

−π3e\frac{-\pi}{3\sqrt{e}}

C

3eπ\frac{3\sqrt{e}}{\pi}

D

−3eπ\frac{-3\sqrt{e}}{\pi}

Answer

−3eπ\frac{-3\sqrt{e}}{\pi}

Explanation

Solution

To find F(π3)F(\frac{\pi}{3}), we first need to evaluate the indefinite integral F(x)=∫ecos⁡x(1+xsin⁡x)x2dxF(x)=\int \frac{e^{\cos x}(1+x\sin x)}{x^2}dx.

Let's examine the integrand: ecos⁡x(1+xsin⁡x)x2\frac{e^{\cos x}(1+x\sin x)}{x^2}. This expression can be rewritten as ecos⁡x(1x2+xsin⁡xx2)=ecos⁡x(1x2+sin⁡xx)e^{\cos x} \left( \frac{1}{x^2} + \frac{x\sin x}{x^2} \right) = e^{\cos x} \left( \frac{1}{x^2} + \frac{\sin x}{x} \right).

Consider the derivative of the function ecos⁡xx\frac{e^{\cos x}}{x} using the quotient rule, ddx(uv)=vu′−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}. Let u=ecos⁡xu = e^{\cos x} and v=xv = x. Then u′=ddx(ecos⁡x)=ecos⁡x(−sin⁡x)u' = \frac{d}{dx}(e^{\cos x}) = e^{\cos x} (-\sin x) and v′=ddx(x)=1v' = \frac{d}{dx}(x) = 1.

So, ddx(ecos⁡xx)=x⋅(ecos⁡x(−sin⁡x))−ecos⁡x⋅1x2=−xsin⁡xecos⁡x−ecos⁡xx2=−ecos⁡x(xsin⁡x+1)x2=−ecos⁡x(1+xsin⁡x)x2\frac{d}{dx} \left( \frac{e^{\cos x}}{x} \right) = \frac{x \cdot (e^{\cos x} (-\sin x)) - e^{\cos x} \cdot 1}{x^2} = \frac{-x \sin x e^{\cos x} - e^{\cos x}}{x^2} = - \frac{e^{\cos x} (x \sin x + 1)}{x^2} = - \frac{e^{\cos x} (1 + x \sin x)}{x^2}

We observe that the integrand given in the problem is exactly the negative of the derivative we just calculated. Therefore, F(x)=∫ecos⁡x(1+xsin⁡x)x2dx=∫−ddx(ecos⁡xx)dxF(x) = \int \frac{e^{\cos x}(1+x\sin x)}{x^2}dx = \int - \frac{d}{dx} \left( \frac{e^{\cos x}}{x} \right) dx F(x)=−ecos⁡xx+CF(x) = - \frac{e^{\cos x}}{x} + C where CC is the constant of integration.

Now, we use the given condition F(π)=−1eπF(\pi)=\frac{-1}{e\pi} to find the value of CC. Substitute x=πx=\pi into the expression for F(x)F(x): F(π)=−ecos⁡ππ+CF(\pi) = - \frac{e^{\cos \pi}}{\pi} + C Since cos⁡π=−1\cos \pi = -1, we have: F(π)=−e−1π+C=−1eπ+CF(\pi) = - \frac{e^{-1}}{\pi} + C = - \frac{1}{e\pi} + C We are given F(π)=−1eπF(\pi) = \frac{-1}{e\pi}. So, −1eπ=−1eπ+C\frac{-1}{e\pi} = - \frac{1}{e\pi} + C This implies C=0C=0.

Thus, the function F(x)F(x) is: F(x)=−ecos⁡xxF(x) = - \frac{e^{\cos x}}{x}

Finally, we need to find F(π3)F(\frac{\pi}{3}). Substitute x=π3x=\frac{\pi}{3} into the expression for F(x)F(x): F(π3)=−ecos⁡(π3)π3F\left(\frac{\pi}{3}\right) = - \frac{e^{\cos (\frac{\pi}{3})}}{\frac{\pi}{3}} Since cos⁡(π3)=12\cos (\frac{\pi}{3}) = \frac{1}{2}, we have: F(π3)=−e1/2π3F\left(\frac{\pi}{3}\right) = - \frac{e^{1/2}}{\frac{\pi}{3}} F(π3)=−eπ3F\left(\frac{\pi}{3}\right) = - \frac{\sqrt{e}}{\frac{\pi}{3}} F(π3)=−3eπF\left(\frac{\pi}{3}\right) = - \frac{3\sqrt{e}}{\pi}