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Question: $\int e^{\sin x}(\frac{x\cos^3x-\sin x}{\cos^2x})dx$...

∫esin⁡x(xcos⁡3x−sin⁡xcos⁡2x)dx\int e^{\sin x}(\frac{x\cos^3x-\sin x}{\cos^2x})dx

Answer

esin⁡x(x−sec⁡x)+Ce^{\sin x} (x - \sec x) + C

Explanation

Solution

The given integral is: I=∫esin⁡x(xcos⁡3x−sin⁡xcos⁡2x)dxI = \int e^{\sin x}\left(\frac{x\cos^3x-\sin x}{\cos^2x}\right)dx First, simplify the expression inside the parenthesis by splitting the fraction: xcos⁡3x−sin⁡xcos⁡2x=xcos⁡3xcos⁡2x−sin⁡xcos⁡2x\frac{x\cos^3x-\sin x}{\cos^2x} = \frac{x\cos^3x}{\cos^2x} - \frac{\sin x}{\cos^2x} =xcos⁡x−sin⁡xcos⁡x⋅1cos⁡x= x\cos x - \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} =xcos⁡x−tan⁡xsec⁡x= x\cos x - \tan x \sec x So the integral becomes: I=∫esin⁡x(xcos⁡x−tan⁡xsec⁡x)dxI = \int e^{\sin x}(x\cos x - \tan x \sec x)dx This integral is of the standard form ∫eg(x)(f′(x)+f(x)g′(x))dx=eg(x)f(x)+C\int e^{g(x)}(f'(x) + f(x)g'(x))dx = e^{g(x)}f(x) + C. In this case, let g(x)=sin⁡xg(x) = \sin x. Then, g′(x)=cos⁡xg'(x) = \cos x. We need to find a function f(x)f(x) such that f′(x)+f(x)g′(x)f'(x) + f(x)g'(x) equals the term in the parenthesis, i.e., f′(x)+f(x)cos⁡x=xcos⁡x−tan⁡xsec⁡xf'(x) + f(x)\cos x = x\cos x - \tan x \sec x.

Let's try to identify f(x)f(x). Consider f(x)=x−sec⁡xf(x) = x - \sec x. Now, let's find f′(x)f'(x): f′(x)=ddx(x−sec⁡x)=1−sec⁡xtan⁡xf'(x) = \frac{d}{dx}(x - \sec x) = 1 - \sec x \tan x Substitute f(x)f(x) and f′(x)f'(x) into the expression f′(x)+f(x)cos⁡xf'(x) + f(x)\cos x: (1−sec⁡xtan⁡x)+(x−sec⁡x)cos⁡x(1 - \sec x \tan x) + (x - \sec x)\cos x =1−sec⁡xtan⁡x+xcos⁡x−sec⁡xcos⁡x= 1 - \sec x \tan x + x\cos x - \sec x \cos x Since sec⁡xcos⁡x=1cos⁡x⋅cos⁡x=1\sec x \cos x = \frac{1}{\cos x} \cdot \cos x = 1, the expression simplifies to: =1−sec⁡xtan⁡x+xcos⁡x−1= 1 - \sec x \tan x + x\cos x - 1 =xcos⁡x−sec⁡xtan⁡x= x\cos x - \sec x \tan x This matches the expression inside the parenthesis in the integral. Therefore, the integral is of the form ∫eg(x)(f′(x)+f(x)g′(x))dx\int e^{g(x)}(f'(x) + f(x)g'(x))dx with g(x)=sin⁡xg(x) = \sin x and f(x)=x−sec⁡xf(x) = x - \sec x. The solution to the integral is eg(x)f(x)+Ce^{g(x)}f(x) + C. I=esin⁡x(x−sec⁡x)+CI = e^{\sin x}(x - \sec x) + C

The final answer is esin⁡x(x−sec⁡x)+C\boxed{e^{\sin x} (x - \sec x) + C}.

Explanation of the solution:

  1. Simplify the integrand: The given integral ∫esin⁡x(xcos⁡3x−sin⁡xcos⁡2x)dx\int e^{\sin x} \left(\frac{x \cos^3 x - \sin x}{\cos^2 x}\right) dx is simplified by splitting the fraction inside the parenthesis: ∫esin⁡x(xcos⁡x−tan⁡xsec⁡x)dx\int e^{\sin x} (x \cos x - \tan x \sec x) dx.
  2. Identify the pattern: This integral matches the form ∫eg(x)(f′(x)+f(x)g′(x))dx=eg(x)f(x)+C\int e^{g(x)} (f'(x) + f(x) g'(x)) dx = e^{g(x)} f(x) + C. Here, g(x)=sin⁡xg(x) = \sin x, so g′(x)=cos⁡xg'(x) = \cos x.
  3. Find f(x)f(x): We need to find a function f(x)f(x) such that f′(x)+f(x)cos⁡x=xcos⁡x−tan⁡xsec⁡xf'(x) + f(x) \cos x = x \cos x - \tan x \sec x. By inspection, we find that f(x)=x−sec⁡xf(x) = x - \sec x.
    • Verify: f′(x)=1−sec⁡xtan⁡xf'(x) = 1 - \sec x \tan x.
    • Then f′(x)+f(x)cos⁡x=(1−sec⁡xtan⁡x)+(x−sec⁡x)cos⁡x=1−sec⁡xtan⁡x+xcos⁡x−1=xcos⁡x−sec⁡xtan⁡xf'(x) + f(x) \cos x = (1 - \sec x \tan x) + (x - \sec x) \cos x = 1 - \sec x \tan x + x \cos x - 1 = x \cos x - \sec x \tan x, which matches the integrand.
  4. Apply the formula: Since f(x)f(x) and g(x)g(x) are identified, the integral evaluates to eg(x)f(x)+C=esin⁡x(x−sec⁡x)+Ce^{g(x)}f(x) + C = e^{\sin x}(x - \sec x) + C.

Answer: esin⁡x(x−sec⁡x)+Ce^{\sin x} (x - \sec x) + C