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I was solving differential equation
$ {x\cos x}\frac {dy}{dx} + y(x\sin x+\cos x)=1$
which on dividing by $x\cos x$ becomes FOLD(first order linear differential) equation.

But I am stuck at following integral. Can anyone help solve this integral? An alternate approach to the problem is also welcome.

$$\int\frac {e^{\cos x}}{\cos x}dx $$

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  • $\begingroup$ That is a good approach, you get a linear equation with an Integrating Factor (IF) that is NOT as you show - check your algebra. Maybe you are incorrectly writing the Integrating Factor. $\endgroup$ – Moo Jul 19 '17 at 4:27
  • $\begingroup$ Yes, I did a calculation error. $\endgroup$ – user458519 Jul 19 '17 at 9:11
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Another approach:

By dividing both sides by $\cos^2 x$ we get

\begin{align*} (x\sec x)\frac{dy}{dx}+(x\tan x\sec x+\sec x)y&=\sec^2 x\\ \frac{d}{dx}\left[(x\sec x)y\right]&=\sec^2 x \end{align*}

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Step 1: multiply by $dx$

Step 2: $(xy \sin x + y \cos x - 1) \ dx + x \cos x \ dy = 0$

$\implies M = (xy \sin x + y \cos x - 1) \ \text{and} \ N = x \cos x$

Step 3: Check if exact, i.e. if $ M_y = N_x $ (Hint, this is not exact)

Step 4: $ \dfrac{M_y - N_x}{N} = 2 \tan x \ \ $ i.e. is a function of x only

Now use integrating factor

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